M1 Momentum with Kinematics | Difficult Questions from Recent Papers | Part 2 | AS/A level Math 9709

Added: 2026-05-09

[00:00:00]Two particles P and Q of masses 4 m kg and 3 m kg respectively they are at rest. They are at rest on a smooth horizontal ground.

[00:00:09]P is projected directly towards Q with speed 6.5 m/s.

[00:00:14]Okay. So we have a horizontal ground.

[00:00:20]There are two particles. One is P and the other is Q. Let's represent them here. P and Q.

[00:00:29]P projected at a speed of 6.5 m/s.

[00:00:34]After the collision, they move in the same direction with speeds 2 and V. Show that V is equal to six. Now, first part is fairly straightforward. This is conservation of momentum. We know about their speeds before collision. Before collision, P is moving at 6.5 m/s. Q is at rest. We can consider the right side positive. If you consider the right side positive then after collision P moves in the same direction. So P and Q move in the same direction. The only possibility is that they move in the right direction. Right?

[00:01:15]they move in the right direction because it's not possible that Q is at rest and P is moving to the right that after collision both of them keep start moving to the right to the left side that's not possible right so both particles are moving to the right after collision after collision we know the speed of P is 2 m/s speed of Q is V apply conservation of momentum we know the masses this mass is 4 m kg and this mass is 3 m kg Momentum before collision that should equal momentum after collision 4 m into 6.5 + 0 momentum after collision 4 m into 2 + 3 m into v. m gets cancelled because that's appearing in every single trip.

[00:02:12]So we get 4 into 6.5 which is 27 right 27 = 8 + 3 * v and that will give the value for v. You can see what that is 19 over3. Is that what we're getting?

[00:02:24]There's something wrong there.

[00:02:27]Uh 4 m and and there we had 3 m 8 + 3 v 27. Let me see if I've copied values correctly. 6.5 there, two there, 26.

[00:02:41]That's a 4 into 6. I I should probably just do that. I 6.5 5 into 4, that's 26.

[00:02:46]Okay, sorry, I wrote that as 27. 26 and then that gives us 18 over 3. There's still some problem with that. Right.

[00:02:56]Uh, I need to check my working again.

[00:03:00]That's what happens when you get distracted in the middle. 4 m 3 m they were at rest initially 6.5 m/s is the velocity of P in the beginning take velocity of Q was zero after collision speed of P is 2 and speed of Q is what we have to figure out okay so then I just input values here 4 M into 6.5 plus momentum of Q that is then 4 M into Sorry velocity equal to 6. Yes sir.

[00:03:37]>> Okay. Yeah. V equal to 6. I don't know what I was thinking. 18 divided by 3 that is 6.

[00:03:45]I had something else in my mind. V6 R.

[00:03:51]That's Yeah. I'm sorry. I was I was doing some other division in my head. 18 I was not getting six. Okay.

[00:04:03]Uh let's move on to the next one now.

[00:04:08]Yes, I'm also tired.

[00:04:13]It's okay.

[00:04:15]Ch. Give that given that the energy lost in the collision between P and Q is 45 JW. Find the value of M.

[00:04:26]Given that the energy lost in the collision is this. Find the value of M.

[00:04:30]Now what's the initial energy? What's the initial energy?

[00:04:35]Initial energy potential energy will be the same, right? Because there's no change in potential energy. It's a horizontal surface. There's only change in kinetic energy. So we that's that's the only energy that we consider for that reason.

[00:04:49]Initial kinetic energy in the very beginning. What is that? / into MV² for P that will be half into 4 M into 6.5^ 2 plus kinetic energy for Q that will be zero. You can simplify this and that will give you 2 into 6.5. I think I should just use a calculator now. 2 into 6.5^ squar or 84.5 84.5 * m then what about the final kinetic energy at the end kinetic energy for P that is/ into 4 m into 2^ 2 +/ into 3 m into V ^ 2 V value is six. You input that there and you'll get some number from this as well. You can see what that is.

[00:05:53]What's this number?

[00:05:56]8 + 54 which is 62. Yeah. 62 * n.

[00:06:07]Now they are saying the energy lost in the collision is 45 jewles energy lost.

[00:06:13]Now what's the difference in these two?

[00:06:18]84.5 * m - 62 m that should equal 45 and that will give you a value for m you can see what that is that turns out to be two right two I would wait for a confirmation is it 2 kg yes no yeah okay m turns out to be 2 kg Okay.

[00:06:49]Okay. There hasn't been a single difficult question in this paper until now.

[00:06:55]Let's look at this. Now, after Q has moved a distance of 9 m. From the point at which P and Q first collided, it hits a vertical wall perpendicular to the direction of motion of Q and rebounds.

[00:07:09]The speed of Q after hitting the wall is 0.5 and distance from the wall to the point at which the second collision between P and Q occurs. Okay, now this looks slightly tricky.

[00:07:26]Okay, so we knew nothing about their distances from the wall earlier.

[00:07:35]Now they're saying after Q has moved a distance of 9 m from the point at which P and Q first collided, it hits a vertical wall perpendicular to the direction of motion of Q and rebounds. Let's make a vertical wall here. So let's let's suppose we've got this vertical wall and they were moving on a horizontal surface earlier.

[00:07:59]The point where they collided with each other earlier, they're saying that point is at a distance of 9 m from that wall.

[00:08:12]Okay, the distance of that wall is 9 m from the point of first collision.

[00:08:22]It rebounds the speed of Q after hitting the wall is 0.5. Okay, so Q rebounds at a speed of 0.5 m/s.

[00:08:30]Remember the surface is smooth, so there's there's no decrease in speed while it is moving on the surface. Okay, find the distance from the wall to the point at which the second collision between P and Q occurs. Okay, it's also not that complicated. If you have already done relative speed questions earlier, this is not the most difficult of those. How do we think about this?

[00:08:58]Now, now while Q moves to the right and reaches that wall, Q would also P would also be moving to the right side.

[00:09:11]Let's figure out at what point P would be when Q collides with the ball. Okay.

[00:09:19]Now what I'll do is I'll mark these points. I I'll call this point something. Let's call this point X and I'll call this point Y for now. X to Y. Let's consider what let's uh consider what we know for this part of this journey. We know displacement that is 9. We're talking about Q, right?

[00:09:45]Q when it goes from X to Y. What do we know about that displacement that is 9 m and the velocity that is constant after that collision? Q was moving at a speed of 6 m/s.

[00:10:00]So velocity is six. We can find the time taken for Q to reach that wall. What's the time taken? It's constant speed, right? When speed is constant, we can just do distance over velocity 9 / 6.

[00:10:12]Time taken for Q to hit the wall that is 1.5 seconds.

[00:10:19]Time taken by P or in fact by Q until it hits the wall.

[00:10:33]Okay, that's the first thing that you figure out. Now if this is the time taken when Q hits the wall, P would have also traveled some distance in that time. How much distance would P have traveled in that time? Let's figure that out.

[00:10:54]I'll do it on the side here.

[00:10:58]Let's call this point something as well in the middle until which P would have moved. Let's suppose this point here was Z. So for P, what's the distance X to Z? We want to figure that out. We know the velocity for P. What was the velocity for P? 2 m/s after that first collision. So velocity was 2 unit 2 m/s. We also know the time taken. Now time is 1.5. Now Z is the point at which P would be when Q hits the wall. Let's find the displacement for this point velocity into time and that gives us 3 m. So that tells us that this distance is 3 m and if that is 3 m then what about the remaining distance from this point to the wall that is 6 m.

[00:11:57]Now, at this point, P is moving to the right at a speed of 2 m/s.

[00:12:03]Q is moving to the left at a speed of 0.5 m/s.

[00:12:08]P is going in this direction. Q is going in this direction. How much time would they take to cover this 6 m distance?

[00:12:16]Let's figure that out.

[00:12:20]The time that we're looking for is the distance from the wall to the point at which second collision takes place.

[00:12:25]We're looking for distance. Okay.

[00:12:33]So now from this point onward we can say the relative speed of the particles what is that going to be? They're both moving in opposite directions. Every second P is going 2 m to the right and Q is going 0.5 m to the left. What's the relative speed going to be? 2 + 0.5 and that turns out to be 2.5 m/s.

[00:13:05]That's the relative speed of the particles.

[00:13:09]Every second they're getting they're getting closer by a distance of two 2.5 m. So that's the relative speed. What's the time taken for this part of the journey? Can we find that time taken?

[00:13:22]6 m divided by 2.5.

[00:13:26]That will give you some value. You can see what this is. 6 over 2.5. What's this number? 2.4.

[00:13:31]2.4 seconds.

[00:13:34]This is the time until the second collision.

[00:13:40]Time until second collision.

[00:13:46]after Q hits the wall.

[00:13:51]This is the time until second collision after P hits the wall. 2.4 seconds. Now that we have figured out this time taken, we are looking for the distance from the wall at which the second collision happens.

[00:14:07]Let's suppose the second collision happens somewhere around here.

[00:14:14]What is the distance traveled by Q1 that time? It speed was 0.5 m/s.

[00:14:23]You can say distance traveled by Q until collision after hitting the ball.

[00:14:46]What would that be? the speed 0.5 into the time taken. Speed note time that will give you the distance 0.5 into 2.4 and that gives you 1.2 m.

[00:14:59]This is where the second collision is going to happen. The distance from the wall of this point whatever this point is let's call this something as well.

[00:15:09]The point of collision this point distance from the wall to this point is 1.2 m. That is what we were looking for.

[00:15:21]Make sense to everyone.

[00:15:24]For relative speed, do we always add the speeds? Do we always add the speeds? No, we don't always add the speeds. It depends on the directions. Let's suppose we have two objects going like this. One is going to the right by 6 m per at 6 m/ second. Another is going to the left by 2 m at 2 m/s.

[00:15:48]Now, if they were 100 m away from each other, every second they're getting closer by 8 m. In this case, we would add those speeds when they're moving in opposite directions.

[00:16:00]If they're moving in the same direction, if A is moving to the right, B is also moving to the right. At what speed are they getting closer to each other? Every second A moves to the right by 9 m. B moves to the right by 4 m.

[00:16:14]By how much distance do they come closer to each other? 9 - 4 and that is 5. In this case, when they're moving in the same direction, the relative speed is going to be the difference between those two values.

[00:16:29]Make sense?

[00:16:33]Any issues with this?

[00:16:37]Okay. Now, let me repeat this part again. This last part again.

[00:16:44]Let me repeat this last part again.

[00:16:50]So, what is it that they were asking about? They asked about the distance from the wall of the point at which the second collision happens. Now look at the different parts of the journey.

[00:17:03]Initially P and Q were both here after the first collision. Now they said Q moves to the right by a distance of 9 m until it hits the wall. Now P would also travel some distance in that time. So firstly what we're trying to figure out is at what point P would be when Q collides with the wall. Now how do you figure that out?

[00:17:27]We are given the distance that Q travels and the speed at which Q travels that distance from that we can find the time taken until collision.

[00:17:40]Okay time taken until it hits the wall.

[00:17:42]1.5 seconds. Now in those 1.5 seconds how much would P have traveled its velocity was 2 time 1.5 it would have traveled 3 m. So now when Q collided with the wall P was at this point here such that the distance between P and Q was only 6 m at this point. Now from this point onward P is moving to the right at 2 m/s. Q is moving to the left at 0.5 m/s.

[00:18:17]When will the collision happen? The collision will happen when this 6 m distance between those two particles is covered.

[00:18:29]Now, how much time would it take for them to cover this distance? One is moving at a speed of 2 m/s.

[00:18:36]Another is moving at a speed of 0.5 m/s.

[00:18:39]So every second they're getting closer by 2.5 m/s. P is coming two to the right. Q is coming 0.5 to the left. We call this relative speed. Relative speed of these particles is 2.5 m/s. Now if the speed is 2.5, distance was 6. We get the time taken for the collision to happen 2.4 seconds. Now in that time how much would have Q traveled? We just multiply the speed of Q 0.5 with that time and that gives us the distance that Q would have traveled and therefore the distance from the wall when the collision happens.

[00:19:21]Is that okay?

[00:19:25]Make sense? Right.

[00:19:30]Yes. So I explained that when they're moving in the opposite direction, the relative speed is going to be their sum.

[00:19:36]When they're moving in the same direction, the relative speed is going to be their difference. Okay. A particle P of mass 0.1 kg is projected vertically upwards. Vertically upwards with speed 30 m/s.

[00:19:52]Mass is 0.1 kg. Speed is 30 m/s.

[00:19:56]At the same instant, a particle Q of mass 0.4 4 kg is projected upwards with speed 10 m/s.

[00:20:05]From a height of 15 m above the ground, P and Q move in the same vertical line.

[00:20:09]Find the height above the ground at which P and Q collide. That's a standard chyntics question. You can make a rough sketch to understand the situation better. P is projected from ground at a speed of 30 m/s.

[00:20:31]Its mass is 0.1 kg.

[00:20:39]Q is projected from a point 50 m above the ground. So let's suppose we've got some building here and Q is projected from there.

[00:20:59]It's already at a height of 15 m above the ground.

[00:21:04]Okay, so Q is projected from here now at 10 m/s.

[00:21:13]They are both in the same vertical line.

[00:21:15]I've represented them on the side as in one on the left, one on the right just to understand the situation better. But they're in in the same vertical line.

[00:21:24]Right? So Q is starting from this point in this direction like this.

[00:21:30]Now at some point they collide. That means they would be at the same height.

[00:21:36]That is when the collision is going to happen. Right? At some point they are going to be at the same height.

[00:21:42]We need to find that height at which the collision happens.

[00:21:47]Let's represent the information that we've got about both the particles P and Q. Both are projected in the upwards direction. It could be two-way motion.

[00:21:57]Since they're projected in the upward direction, they might come downwards at some point.

[00:22:03]We will just consider upwards positive for both of them. Now, what about initial velocities? We know the initial velocities. Initial velocity for P is 30. For Q, it is 10. What else do we know? We know acceleration. Acceleration is going to be -10, right? We're taking upwards positive. Gravity acts in the downwards direction. Gravity acts downwards.

[00:22:26]So, acceleration is is going to be -10 for both of them. What else? Is there anything else that we know about? Do we know final velocities at that point? No.

[00:22:41]Do we know f? Do we know time taken? No.

[00:22:44]Displacement. We have some information about displacement.

[00:22:48]What do we do now?

[00:22:51]What we think about what is common between them. Is there anything that is common between them? They both started from the same time. They started from the same time and until they reach this height, they would have spent the same amount of time until that point. Right?

[00:23:07]they would have spent the same amount of time time until that point.

[00:23:12]So the time is common from for them. If we considered the time of P to be T, the time for Q is also going to be T. So time is common between them.

[00:23:22]And we know something about displacement. What do we know about displacement? At the point of collision, we could say P would have traveled a distance which is SP and Q would have traveled a distance which is SQ. We have a relation between SP and SQ. We can say SP is equal to SQ + 15.

[00:23:51]We can make an equation like this. Solve this equation and find the value of t from this.

[00:23:57]All right. S is equal to sq + 15.

[00:24:03]Now, which equation connects these things? U a t and s. S= U +/ A T². Input values in that.

[00:24:11]For P, it will be 30 into T +/ into -10 into T ^ 2. That equals 10 into t +/ into -10 into t ^ 2 + 15.

[00:24:27]Simplify this you get 30t - 5t ^ 2 = 10 t - 5t ^2 + 15. 5t ^2 and 5t ^2 they will get cancelled and we're left with 20t = 15. The value of t turns out to be 0.75.

[00:24:47]This is the time after which the collision happens.

[00:24:51]Now what's the distance above the ground? The height above the ground at which the collision happens.

[00:24:58]SP will be the height above the ground.

[00:25:00]We can find the value of SP. Now let's see what that turns out to be. Distance traveled uh displacement of P from ground. that is given by this 30 into t that is 0.75 +/ into a which is -10 into 0.75 squared that will be the height above the ground. You can see what that is.

[00:25:26]What's this number?

[00:25:34]Has anyone done this?

[00:25:44]19.6875.

[00:25:45]Thank you. 19.6875.

[00:25:50]Round it up to three significant figures and that will be 19.7.

[00:25:55]That's the distance that we were looking for. That's the height above the ground.

[00:26:00]Okay.

[00:26:04]Any issues with this? If it's an exact number, you just leave it as it is. Is it is it exact or is it rounded off 6875?

[00:26:16]Okay. So, if it's exact, you leave it as it is. Then you don't need to round it off. You leave it as it is. If there was a large number of decimals which were not ending, then you round off to three significant figures. If it's exact, you leave it as it is. Or a or as a fraction, that will also be fine.

[00:26:36]Okay, that was part A. Now they're saying when P and Q will write the coal S, find the speed of the combined particle at the instant it reaches the ground. Okay, now this is a complicated part. It looks like it when we have chyntics and momentum together, they become slightly more complicated. Let's think about this. Now I'll just copy that diagram from there.

[00:27:10]Okay. So this is what the situation was.

[00:27:12]Now I'll remove irrelevant information from there. So we don't care what happened before. Uh all we care about is at the point the collision is happening.

[00:27:23]Right? So actually just ignore the rest of it.

[00:27:28]Let's say that's the point at which the collision is happening.

[00:27:32]The height of that point all of that the height where a collision happens we found that in the first part that was 19.6675 what's the fraction value for that 19.675 6875 315 over 16. Okay. So this was 315 / 16 that you found in the previous part.

[00:27:58]So we could say this height is 315 / 16 m above the ground at which collision happens.

[00:28:08]Now what are their speeds at which the collision happens?

[00:28:16]At what speed uh what are the speeds of both the particles? We would have to find that because we need to apply conservation of momentum. Can we find their speeds before collision?

[00:28:26]Let's do that.

[00:28:29]Speeds before collision. VP equals U + 80 and VQ= U + 80.

[00:28:38]Initial velocity for P was 30. So 30 + - 10 into 0.75 that will give you 22.75 22.25 not 22.5 yeah and then VQ it started at 10 so + - 10 into 0.75 and that will give you 2.5 m/s. So at the point collision happens, P is moving at 22.5 m/s in the upwards direction and Q is moving at 2.5 m/s in the upwards direction. So let's say this is P and this is Q. For Q, we know it is moving upwards at 2.5 m/s. And for P, we know it is moving at 22.5 m/s.

[00:29:40]That's before collision. They're both moving in the upwards direction. Can somebody confirm these calculations as well, please? They're both positive, right? And these are the numbers.

[00:29:55]Yes. Considering both of them positive in the upwards direction these are the numbers that you get right.

[00:30:03]Okay. Now we know before collision and after collision situation for both the particles before and after we have particle P and particle Q.

[00:30:16]Let's consider upwards direction as positive. Before collision we know velocity of P is 22.5 and velocity of Q is 2.5. We know their masses as well.

[00:30:26]Mass of P was 0.1 and Q was 0.4 0.1 kg and 0.4 kg.

[00:30:36]After collision what happens is they coales and become one single object. So after collision now they become one single object PQ. Let's say the combined particle.

[00:30:47]Mass of this combined particle is going to be 0.5 kg. What about the speed of this combined particle? We can find that.

[00:30:55]Apply conservation of momentum. Momentum before collision that equals momentum after collision.

[00:31:02]What's the momentum before collision?

[00:31:03]0.1 into 22.5 plus 0.4 into 2.5.

[00:31:12]That equals 0.5 into BQ. And that will give you some value for VQ. You can see what that is.

[00:31:20]What's this number?

[00:31:27]2.25 + 1 divided by.5 6.5 VQ turns out to be 6.5 m/s. This is a positive number. Since it's a positive number, what that means is we took upwards positive.

[00:31:52]After collision, the combined particle is going to start moving in the upwards direction. Okay, it's going to start moving in the upwards direction. So now the situation is the collision has happened. Okay, after this collision, after collision, the initial velocity of the combined particle, we have figured that out. That is 6.5.

[00:32:18]And in which direction is the particle moving? It is moving in the upwards direction since it's a positive number.

[00:32:25]So we can say the the combined particle now it's going to go up. It journey is going to look like this. uh it's going to go up for a while and then it's going to go down after that.

[00:32:37]Okay, this is ground. What information do we have about this particle? Now we can consider upwards positive. It's two-way motion. When it's two-way motion and vertical direction, we consider upwards positive. Initial velocity, what's that?

[00:32:53]6.5. And that's in the upward direction.

[00:32:56]So, we consider this positive.

[00:32:58]What else do we know? Acceleration if you're considering upwards positive acceleration is going to be -10. Is there any other information that we've got?

[00:33:08]We know displacement. We know the point at which collision happened that was 315 over 16.

[00:33:18]The point at which collision happened that was 35 / 16.

[00:33:30]Okay, this is 315 / 16.

[00:33:36]So what's the value of displacement that we take here? Do we say 315 over 16? Is that the displacement value?

[00:33:46]3 and 5 16. Will that be correct?

[00:33:50]Will this be correct or not?

[00:33:54]It'll be negative. We're considering upwards positive. We're considering upwards positive. This is the starting point. End point is down there.

[00:34:02]Displacement is to the in in the downward direction. So the displacement value that we take is 315 /6. And now what are we looking for? Speed of the combined particle when it reaches the ground. The final speed that's what we're looking for. You can see which equation connects these four things.

[00:34:19]v ^2 = u ^2 + 2 a s. Input values in that and that will give you the result.

[00:34:25]V ^2 = 6.5^ 2 + 2 into a which is -10 into s which is -315 / 16. Input values in a calculator. See what value of v you get from that. you'll get plus minus it will just take positive value because we're just considering the speed. Okay.

[00:34:48]6.5^ 2 + 2 into -10 into -315 / 16 and that will give you the result that's 436 and then square root of that.

[00:35:04]Do we want an exact answer? No.

[00:35:07]So the speed turns out to be 20.88 to three significant figures that will be 20.9 m/s and that's our final answer.

[00:35:23]Is that all right? Why is displacement negative? We talked about this yesterday as well.

[00:35:31]This is the starting point.

[00:35:34]This is the end point.

[00:35:37]Start end we are considering upwards positive. In which direction is displacement?

[00:35:45]From the starting point to the end point. Displacement is is in the downward direction.

[00:35:51]End point is below. Starting point is above.

[00:35:55]So overall displacement is in the downward direction. We considered upwards positive. So displacement is going to be negative. The part the second part. Now this seems to be difficult. Whenever you have chyntics and momentum collisions happening, these questions tend to be among the most difficult. Let's see if that's the case with this question or not. Three particles P, Q and R, they are of mass are 0.6 kg,4 kg,8 kg. They're at rest.

[00:36:22]Distance from P to Q is 3 m. Okay. Uh and the distance from Q to R is also 3 m. P is projected directly towards Q with speed 3 m/s.

[00:36:37]P is projected towards Q with speed 3 m/s.

[00:36:41]After they collide, P continues to move in the same direction with speed 1.5 m/s. Find the speed of Q after collision. Okay, the first part is straightforward. Just a simple application of Newton's of uh conservation of momentum. So you write down what we know about before collision, what we know about after collision. We've got two particles, one is P and the other is Q.

[00:37:02]Considering the right side positive, P is moving at 3 m/s to the right. Q is at rest at that point. After a collision, P continues in the same direction at 1.5 m/s. We are looking for the velocity of Q.

[00:37:19]We have their masses as well. This is 0.6 kg and this is 0.4 kg.

[00:37:26]Apply conservation of momentum. Momentum before collision that should equal momentum after collision.

[00:37:35]0.6 into velocity of P plus 0.4 into 0. On the other side we have 0.6 into 1.5 + 0.4 into VQ and that will give you the value for VQ. You can see what that is 1.8 and then minus this from the other side. 6 into 1.5 is 0.9 that equals 0.4 VQ.9 over.4 four. What's that? 9 / 4.

[00:38:02]That should be 2.25.

[00:38:05]That's the velocity that we get. Okay, that's the first part. This is all that we had to use. Okay, let's move on. Part two. Now, it says in the subsequent collision between Q and R, these particles coales. Find the speed of the combined particle after this collision.

[00:38:20]That's also straightforward. U we have another collision now. Before the second collision, what do we know about both the particles? Before the second collision Q and R, at what velocities are they moving after the first collision, P is moving at 2.25. So that's the before collision speed. For the second collision, right? 2.25. R is at rest. So that is 0 m/s.

[00:38:50]Afterwards what happens is they coalesce and become a single object. Let's call that object have they named it something?

[00:39:00]Let's call it C the combined particle.

[00:39:06]We have to find the speed of this particle. Now what about the masses? Q and R were 4 and8 respectively. So this is8 kg. This is4 kg. So the total mass what is that going to be? That will be 1.2 2 kg sum of these two masses. Apply conservation of momentum. Momentum before collision 0.8 into 2.25 plus the momentum of R that is 0 equals momentum after collision 1.2 into velocity of the combined particle. And you can see what this turns out to be.8 into 2.25 divided by 1.2. What's that number?

[00:39:46]What's this number anyone?

[00:39:50]75. This is the velocity that we get for the particle after collision for the combined particle after collision 75.

[00:40:05]Okay, that's our final answer. Let's move on. Okay, the first two parts were fairly straightforward. Now they're saying find the time that it takes from when P is initially projected until the instant at which P collides with the combined particle.

[00:40:23]Okay. Now time when from when P is initially projected until when P collides with the combined particle. So we've got two collisions already and then they're saying P collides with the combined particle again after that. What's the total time that it takes?

[00:40:46]Now uh for this we need to know a concept that some of you might not have done.

[00:41:02]It's just simple logic but it helps to just do it uh in a bit more in a bit of detail right now because we will need that for later papers as well. So let me actually go through that quickly in recent papers we have to use that quite a lot now.

[00:41:19]So what I've started doing is I've started teaching that as a standard thing now recently let me go through that the concept of relative speed. Now it's simple logic but let's just make sure that we understand it well. Let's suppose let me actually hide this for now.

[00:41:44]There's a particle A that's moving to the right at a speed of 6 m/s. Okay.

[00:41:49]There's another particle B that's moving to the left at a speed of 2 m/s.

[00:41:55]The distance between them initially is 100 m.

[00:42:00]Okay, that's the distance between the two particles initially at the in the beginning 100 m. Now, what's the time taken until collision happens until they collide with each other? Now, the assumption is they're moving at constant velocities, right? If they were moving at different veloc if the velocity was changing, if there was some acceleration or deceleration, then we would use our standard method. We would make some equation for a make some equation for b find their displacements add them up and equate that to 100 like we do in chynatics something like this right make an equation not like this but make an equation for displacement using s= u t +/ a² make an equation for displacement of b as well similarly then you say s a plus sb equal to 100 that's the general method for such problems but when it is constant speed. When it is constant speed, you can do it quickly in one single step. How exactly would you do that? If they're moving at constant speed, what's the time taken until collision? Think of it this way. Every second, how much distance is this particle A traveling? In 1 second, how much distance would A travel? It would travel a distance of 6 m to the right in 1 second. Similarly, B would travel a distance of 2 m to the left in 1 second.

[00:43:26]So in every second, A travels a distance of 6 m to the right, B travels a distance of 2 m to the left. Now out of this 100 m, how much distance is covered in 1 second?

[00:43:43]How much distance have they covered out of this 100 m in 1 second? How much what's the distance now between them?

[00:43:53]92 m. Right? They have come closer by 8 m because six is uh A is moving 6 m right. This is moving 2 m left. So they have moved together at uh by 8 m. This is something that we call relative speed. The speed of the particles relative to each other.

[00:44:15]And in this case we can calculate it like this. We could say let me bring that up from here. Now try that down here.

[00:44:42]Oh What's the relative speed going to be then? We say the relative speed of the particles is 6 + 2 which is 8 m/s.

[00:45:00]That's the speed at which the two particles come closer to each other.

[00:45:06]Okay. So then to find the time until collision all we have to do is distance over the speed that's 12.5 seconds and we would not have to make lengthy equations like these. We could do it directly in one single step 100 / 8 like this. That's the idea of relative speed. Okay. Similarly if they were moving in different directions like this for instance. So, A was moving to the right at 9 m/s. B was moving to the right at 4 m/s.

[00:45:44]How much time does it take for them to collide? Again, use the idea of relative speed.

[00:45:51]How much do they get closer to each other in 1 second? A goes to the right by 9 m. B goes to the right by 4 m. How much closer do they get? 5 m. So they out of this 100 m they cover a distance of 5 m. So the relative speed in this case is going to be 9 - 4 which is 5.

[00:46:14]When particles are moving in the same direction the relative speed is going to be their difference or the difference of their individual speeds. When they're moving in the opposite direction the relative speed is going to be the sum of their individual speeds. Okay?

[00:46:30]Now keep that in mind.

[00:46:34]when we look at this problem now.

[00:46:38]So what was the situation? I'll just copy this diagram here.

[00:46:43]This is the initial situation of the particles P, Q and R. There are three collisions that are happening. There are three collisions that are happening.

[00:46:56]Let's see what's the time taken by all of those three uh collisions. This is the situation.

[00:47:06]Now what about the first collision? So initially we know P started moving to the right at a speed of 3 m/s. Right?

[00:47:14]Now the distance is 3 m. So what's the time until first collision? Time until first collision. We can find that.

[00:47:22]Right?

[00:47:24]time until first collision.

[00:47:28]What would that be? Distance over speed.

[00:47:30]That is 3 over3 and that is 1 second. So until the first collision 1 second passes. Okay.

[00:47:42]Now P has reached this point after 1 second.

[00:47:46]Remember P is here at time= 1 and its speed at time= 1 is 3 m/s. At this point when P and Q collide Q starts moving to the right side. Now Q starts moving to the right side.

[00:48:04]Let me represent that here.

[00:48:08]Q starts moving to the right. Now Q moves to the right at a speed of what did we find there? 2.25. 25. Now Q is moving to the right at 2.25. R is at rest. R is at rest. So how much time does it take to reach R? Now what's the time taken for the second collision? You can say time between first and second collisions. What's that time going to be?

[00:48:38]That's again straightforward. Distance over speed which is 3 over2.25. You can see what that is. What's this number?

[00:48:47]3 over 2.25 1.33 33 33 3. Okay, that's 4 over 3. 4 over 3. That's the time that we have between first and second collisions.

[00:49:02]Now at the end, what happens is we have a combined particle. So now Q and R coales, they become this combined particle. And that combined particle is now moving to the right at a speed of 0.75 m/s.

[00:49:20]And now the third collision has to happen. Now the third collision has to happen between particle P and particle C.

[00:49:32]Okay. Now what's the relative speed going to be?

[00:49:36]Remember what speed is P moving at?

[00:49:39]After the first collision, P was moving at a speed of 1.5 m/s. So that is still traveling at a at a speed of 1.5 m/s to the right side.

[00:49:53]It's traveling at a speed of 1.5 m/s.

[00:49:57]So we can find the relative speed now 1.5 minus 0.75.

[00:50:03]But what about the distance value? What what will be the distance that we take?

[00:50:09]And that do we say P and C are going to be at a distance of 3 m from each other? Do we use that distance or is it going to be something else?

[00:50:27]After the second collision, are P and C going to be 3 m away from each other?

[00:50:35]No, because P continued moving in the right direction.

[00:50:41]P continued moving in the right direction while Q was moving towards R earlier. So in these 4 over 3 seconds that passed between the first and second collisions, P has also traveled some distance to the right side already. P was also moving in that time. Right? So in those 43 seconds when the second collision while the second collision happened P has already moved some distance. So we can say distance moved by P between first and second collisions.

[00:51:19]What would that be?

[00:51:22]How much distance does P speed travel in that time? Its speed was 1.5 m/s. It travels a distance of four.

[00:51:31]It travels it takes 4 over 3 seconds. So 1.5 into 4 over 3. What is that? That is 2 m. Yeah, that's 2 m. That's the distance traveled by P between first and second collisions.

[00:51:48]So P is already somewhere here. Now P has moved to the right and it has come to this point somewhere. Let's say such that this distance traveled by P while Q was going there this is already 2 m when this third collision happens and now at this instant what happens is P is moving to the right at a speed of 1.5 m/s.

[00:52:23]Q uh C the combined particle is moving to the right at 0.75 m/s. The distance between them is now only 1 m out of that 3 m we have 1 m left now. Now now what's the relative speed in this part? Now in the last part relative speed of P with respect to C what would that be?

[00:52:53]1.5 minus 0.75 and that turns out to be 0.75 m/ second.

[00:53:01]What's the distance that it has to covered?

[00:53:05]Uh or in fact time is what we're looking for.

[00:53:08]Time between second and third collisions. We can find that now.

[00:53:18]time between second and third collisions distance 1 divided by the speed and that will give you the time taken for this part 1 / 0.75 what's that number that's also 4 over 3 and that's it now we can find the total time taken find the total [clears throat] time taken see what that turns out to be 1 + 4 over 3 + 4 over 3 and that gives us 11 / 3.

[00:53:55]Two particles P and Q of masses 2 m kg and m kg they are held at rest in the same vertical line. The heights of P and Q above horizontal ground at 1 m and 2 m respectively. P is projected vertically upwards with speed 2 m/s. At the same instant Q is released from rest. Find the speed of each particle immediately before they collide. Okay, let's make a rough sketch to understand the situation better. We've got a horizontal surface and there is one particle that is 1 m above.

[00:54:34]Let's say that that is here. That particle is uh P.

[00:54:40]There's another particle 2 m above that particle is cube.

[00:54:48]Now this is 1 m. This is another 1 m.

[00:54:54]Q starts from rest. It is released from rest and p is projected vertically upwards with speed 2 m/s at the same instant.

[00:55:08]We have their masses given 2 m kg and m kg.

[00:55:19]Find the speed of each particle immediately before they collide.

[00:55:23]Okay. Now what is that what does their journey look like?

[00:55:27]P is moving upwards. Q is moving downwards.

[00:55:31]At some point they're going to collide with each other. Let's suppose that point is here. Let's call that point something. Let's say this this is X.

[00:55:38]Let's write down the information that we've got for both the particles for P and for Q.

[00:55:47]P is moving in the upwards direction. So we can take upwards positive for that. Q is moving in the downward direction. We can take downwards positive for that.

[00:55:54]For P we know for P we know U is equal to 2.

[00:56:05]Acceleration is -10. Gravity acts downwards.

[00:56:12]We're taking upwards positive. So, A will be - 10 for this.

[00:56:16]Is there anything else that we know?

[00:56:19]There's nothing else that we know, right?

[00:56:22]What about Q? For Q, we know initial velocity is zero. What about acceleration?

[00:56:28]Gravity acts downwards.

[00:56:31]Mass is not required here, right? In chynetics when we're applying new uh applying constant acceleration formulas mass does not have anything to do with them right we will have to use that when we use conservation of momentum later on because Q is released from rest so it is moving in the downward direction right you can't release something something from rest and it starts moving in the upward direction that does not happen so Q is going to move in the downward direction right when it is released That's the only possible direction.

[00:57:09]Okay.

[00:57:11]What about acceleration for Q?

[00:57:13]Acceleration is going to be positive 10 because we're considering downwards positive. Gravity also acts downwards.

[00:57:19]So for Q A is going to be positive 10.

[00:57:23]Now we think of what else we can connect them with. Is there anything common between them? Is time common?

[00:57:32]Time is common, right? Because they both start from the same instant and until the point of collision, the same time of time would have passed. Until the time of collision, the same amount of time would have passed. So time is common between them. So if TP is T, then TQ is also going to be T. Okay. Now we also know something about displacement. The distance between P and Q was 1 m. What that means is until the point of collision if P travels SP then Q would have traveled SQ in the downward direction. We cannot assume that their displacements are going to be equal.

[00:58:20]Displacements are not equal. This is this is this will be wrong assumption that whatever distance P moves in the upwards direction Q also moves the same distance in the downward direction. All we know is that the sum of SP and SQ that is equal to 1.

[00:58:39]This is all that we know. SP plus SQ is equal to 1. This is all that we know.

[00:58:47]Now we can make an equation using that.

[00:58:50]SP plus SQ is equal to 1.

[00:58:52]Which equation connects these things? S= UT +/ A T². Let's apply that on P. That will become 2 into T +/ into - 10 into T ^ 2. And then Q 0 into T +/ into 10 T ^ 2 that equals 1.

[00:59:16]Now let's simplify this. It becomes 2t - 5t ^2 + 5t ^2 = 1. The value of t turns out to be 0.5.

[00:59:28]This is the time taken until collision.

[00:59:34]Okay. Now what's the speed of each particle immediately before they collide? We can we can find final speed now, right? V= U + 80. Apply that on P as well as in Q. VP is going to equal U + 80. That means 2 + - 10 * 0.5. And that will give you - 3 m/s.

[00:59:58]VQ U is 0 + a which is 10 into T which is 0.5 and that will give you 0.5 into 10 which is 5 m/s.

[01:00:10]That's the velocity of Q. Now they wanted us to find the speed of each particle.

[01:00:18]If we write -3 we lose a mark. We would say speed of P is equal to 3 m/s and speed of Q that is equal to 5 m/s. That's the first part that we're looking that we wanted to solve. Fairly straightforward standard chyntics question.

[01:00:44]This is what we had to do. Any problems?

[01:00:54]No.

[01:00:57]Now there's an important thing to note here. For P, we took upwards positive.

[01:01:03]We took upwards positive for P.

[01:01:06]But its velocity is turning out to be -3.

[01:01:11]What does that mean? At the point of collision, P is moving in the downward direction.

[01:01:20]So what what has h what is happening here is P was projected in the upwards direction before it collided with Q. It had already started its return journey. It had started returning in the down to to to to the to the ground.

[01:01:39]Q was moving in the downward direction from the top.

[01:01:43]The collision happens when both of them are moving in the downward direction.

[01:01:55]Is that okay?

[01:02:03]So, how is this still equal to one?

[01:02:07]because we're talking about displacement, right?

[01:02:11]So displacement doesn't care about what happened in the middle. All we care about is at the point of collision.

[01:02:22]P was this much above its starting point and Q was this much below its starting point. The sum of displacements is still still going to be one. Yes, the distance traveled for P is going to be greater.

[01:02:36]But that's not what we have in this equation. We have displacement in this.

[01:02:41]Displacement only cares about the current position of the particle.

[01:02:46]Okay.

[01:02:52]So at the point of collision but they are both moving in the downward direction. That is important to note from this result.

[01:03:00]Okay. Okay. Now in the next part they're saying it it's given that immediately after the collision the downward speed of Q is 3.5 m/s. Find the speed of P at the instant that it reaches the ground.

[01:03:12]So now at the point of collision we apply conservation of momentum. First of all Q was above right that was also moving in the downward direction.

[01:03:23]P was below that it's also moving in the downward direction.

[01:03:28]We know the speeds for both of them.

[01:03:32]Q was moving in the downward direction at 5 m/s at the point of collision and P was moving in the downward direction at 3 m/s.

[01:03:45]Let's sum write this information on the side before collision and after collision P and Q.

[01:03:55]Now we have to take one direction positive for all the objects when we're applying conservation of momentum. In chyntics you see I took upwards positive for P downwards positive for Q. That's what you should do for chyntics. The direction in which the particles are moving take that direction positive. But in momentum we have to take the same direction positive for both objects. We can take downwards positive since both of them are moving downwards.

[01:04:23]Velocity of P is three. Velocity of Q is five.

[01:04:28]Before collision, after collision, Q is moving at 3.5 m/s. After collision, what's the speed of P?

[01:04:36]Let's call that VP. Apply conservation of momentum. Momentum before collision, that should equal momentum after collision.

[01:04:44]We are given the mass of P was 2 m kilg, right?

[01:04:51]And mass of Q was m kg. Apply conservation of momentum. Momentum of P before collision that will be 2 M into 3 plus momentum of Q which is M * 5 that equals momentum after collision of P 2 MV VP + M into 3.5. M is appearing in every single term. It will get cancelled out and we're left with 6 + 5 which is 11 on the left side. On the right, we've got 2 VP + 3.5. And that will give you a value for VP. You can see what that is. 7.5 / 2, which is 3.

[01:05:32]No.

[01:05:37]Is there something wrong there?

[01:05:45]3.7 something, right? 3.75. Is that what you're getting?

[01:05:50]keep okay VP is 3.75 m/s and that's the [clears throat] value of velocity of P that we're getting.

[01:06:04]We need to find the speed of P at the instant it reaches the ground. So now what happens after collision?

[01:06:11][clears throat] After collision now P is at some point it is moving in the downward direction.

[01:06:18]at a speed of 3.75 m/s and it hits the ground at some point.

[01:06:27]Now, what do we know about this journey for P after collision?

[01:06:35]Again, we're taking downwards positive.

[01:06:37]Initial velocity is 3.75. Since we're taking downwards positive, acceleration is 10.

[01:06:45]We want to find the final speed time. We don't know what the time is. We don't know what displacement is. But can we find the value for displacement somehow?

[01:06:58]Is there a way to find the value of displacement?

[01:07:05]The collision happened at the point X.

[01:07:08]At this point X, that's where the collision happened, right?

[01:07:15]What's the height above ground at the point of collision? 1 m this and then SP more displacement of P more 1 + SP.

[01:07:26]What we can do is we can find SP before collision first of all.

[01:07:33]Find SP first of all before collision that's this number right? 2t - 5t ^ 2 let's find that value sp before collision that is equal to 2t - 5t ^ 2 right we did that here 2 into t the value for t was 3.75.

[01:08:13]No, the value for t was 0.5.

[01:08:20]2 into 0.5 + 1 /2 into - 10 into 0.5^ 2 that will give you something. What's this number?

[01:08:33]1.25. 25.

[01:08:44]How can it be 1.25?

[01:08:49]0.25. Okay. Before collision, before collision, what was uh the situation was like this.

[01:09:01]P started from here.

[01:09:05]This distance was already 1 m. Q was here. Q was coming downwards.

[01:09:14]P was projected in the upwards direction.

[01:09:18]And then at some point collision happened.

[01:09:22]When we find the value for SP before collision, this value is turning out to be0.25.

[01:09:33]Now what does that mean? If this value is0.25 we considered upwards positive for P there. If the if this displacement is negative what is happening is the collision is actually happening down there like this.

[01:09:56]Q is coming down from there.

[01:10:00]P goes up for a while then it starts going downwards.

[01:10:06]This is the point where they collide with each other. Where P is below its starting point. So Q was coming from there.

[01:10:16]This is where they collide with each other such that from the initial position from the initial position it is 0.25 25 m below this is where the collision is happening. Now the earlier diagram was based on the information that we had available there. We just made a sketch of this that okay this is what it might look like that's that was just for understanding the question but our calculation is telling us the collision is not happening here. The collision is happening at some point when P has returned down there somewhere.

[01:10:59]That does not change anything before.

[01:11:02]All that changes now is since the collision is happening below that initial point of projection for P.

[01:11:12]The height at the point of collision, what is that going to be?

[01:11:19]This height at the point of collision is 0.75.

[01:11:27]This is 0.75.

[01:11:30]So the displacement that we would use here would be 0.75 m. It will be 75 m.

[01:11:43]Okay.

[01:11:45]So let's find the final speed now. V ^2 = U2 + 2 A S input values in that V ^ 2 = 3.75^ 2 + 2 into 10 into 0.75 and that will give you some value for B.

[01:12:04]I think I fish I should be doing calculations myself as well + 2 into 10 into 75 square root of this that's 5.39 5.39 m/s that's the speed of P when it hits the ground this is why I keep stressing it is extremely important that you clearly write down what your positive direction is and then accordingly you should be writing the correct signs for acceleration, displacement, time. Do not just change signs randomly. This displacement was turning out to be negative. It had a meaning. It meant that the collision point was below the starting point.

[01:13:02]That's what I've shown here. Mosa P was projected upwards.

[01:13:08]In the previous part, we already found at the point of collision P was moving in the downward direction.

[01:13:19]So P went in the upward direction for a while and then it started moving in the downward direction. At the point the collision happened, P was below its initial position. So this yellow highlighted point, P is below its initial position. So the point of collision is below the initial position.

[01:13:46]It was at a height of 1 m earlier. Collision happens 0.25 m below that point. So how much does it still have to travel to reach ground after collision?

[01:14:01]It has to travel 75 m below in the uh after collision to reach ground.

[01:14:08]That is why we take displacement value as 0.25 there 75 there.

[01:14:36]Okay. So, Q's displacement will be 1.25. Now, Q displacement 1.25. Okay. Because for Q we were taking downwards positive. So, Q has traveled a distance of 0.2 25 1.25 in the downward direction. For P, the displacement is0.25.

[01:15:01]Add up both of them. 1.25 minus 0 minus 0.25. That gives you one.

[01:15:07]So, the sum of displacements is still one. As long as you're taking your signs correct, the sum is still going to turn out to be one.

[01:15:18]Okay.

[01:15:26]Why did we not take initial velocity of P as neg3 before collision in conservation of momentum?

[01:15:37]Let me repeat that in kyometics we can take different direction positive for different objects and that is recommended that will be more intuitive.

[01:15:49]P was moving in the upwards direction. P was projected in the upwards direction.

[01:15:53]So we can say it's two-way motion. In that case, we take upwards positive.

[01:15:56]Right? So we took upwards positive. For P initially for Q, Q was released and it was moving in the downward direction. So for Q, for Q it took downwards positive.

[01:16:10]From this we got that this velocity of P was -3. Upwards was positive. -3 means P is moving at 3 m/s in the downward direction at the point of collision.

[01:16:26]For Q we took downwards positive and the velocity value was also positive. So for Q we could say it is moving at 5 m/s in the downward direction.

[01:16:39]When we applied conservation of momentum, we took downwards positive for both objects. Because in conservation of momentum, we have to take the same direction positive for all objects.

[01:16:51]Since we're inputting everything in a single equation, we can't take different directions positive when we when applying conservation of momentum. So here we're taking downwards positive for both of them.

[01:17:02]Both of them are moving downwards. So their velocities are positive. A particle A is projected vertically upwards from a point O with a speed of 80 m/s.

[01:17:17]Okay. 1 second later, a second particle B with the same mass as A is projected vertically upwards from O with a speed of 100 m/s.

[01:17:31]So we're projecting this from O at a speed of 80 m/s.

[01:17:39]This particle is A.

[01:17:46]Another particle is projected from the same point and the same mass.

[01:17:54]That's particle B. And it is projected at 100 m/s. This is projected 1 second later.

[01:18:05]Okay. At time t seconds after the first particle is projected, the two particles collide and coales to form a particle C.

[01:18:12]Show that T is equal to 3.5. The time at which collision happens. Now let's say there is a point somewhere at which the two particles are going to collide with each other.

[01:18:23]Let's say this is that point.

[01:18:26]Now the displacement until that point the displacement until that point is going to be the same for both of them.

[01:18:34]When they collide until that point A would have traveled the same amount of distance as B.

[01:18:47]Okay. Now let's write down what information we've got about both the particles about A and B.

[01:18:53]They're both projected in the upwards direction, right? So we can take upwards positive for both of them.

[01:19:00]For a we know the initial velocity that is 80.

[01:19:04]We know acceleration we are taking upwards positive. So acceleration is going to be -10.

[01:19:10]And that's it. This is all the information that we've got for A. For B we know initial velocity is 100 and acceleration is -10.

[01:19:21]Is there anything that we anything else that we can connect the two objects by the two particles by? Is there anything that we can connect them by? Time.

[01:19:29]Right? We know something about time.

[01:19:34]If we say that time taken for a is t. What about the time taken for b? What would that be? Time for b.

[01:19:50]B is projected later. The particle that starts later, it takes less time. Right?

[01:19:56]The particle that starts later, it takes less time.

[01:20:01]For B, it is going to be T minus one.

[01:20:05]That's a standard rule for time delay problems, right? The particle that starts later, it takes less time. The particle that starts earlier, it takes more time.

[01:20:15]Right? I'll just explain that again. So you can think of numbers. For example, if a starts at time equal to0 and reaches that point at time= 10.

[01:20:27]Timeals 10 is let's say the point of collision.

[01:20:30]Now B starts 1 second later. Let's say it starts at t= 1.

[01:20:35]It reaches that point at time= 10 as well. A takes 10 seconds. How much time B? How how much time does B take? 9 seconds. So B takes less time. If the end point is the same and the starting point for B is later, it would take less time. Okay. All right. So we'll have to consider that time as T minus one. Now we know something about displacements.

[01:21:00]At the at the point of collision, the displacements for both the particles are going to be equal to each other. SA equals SB. And that gives us UT +/ A square. Right? That's the equation that we can use. U is 80 into T +/ into A which is -10 into T ^ 2 that is equal to 100 into T -1 + 1 /2 into A which is -10 into t -1^².

[01:21:31]Now we can solve this equation find the value of t from this. The first one gives us the left side gives us 80 - 5t ^ 2. The right side when we simplify this we get 100 t minus 100 can do some cancellation there it becomes min -5 - 5 into t ^ 2 - 2t + 1 simplify this you get 80 t - 5t ^ 2 that equ= 100 t - 100 - 5 t ^ 2 + 10 * t - 5 5 t ^ 2 and 5t ² they get cancelled on both sides.

[01:22:16]On the right side we have 110 t that goes to the other side - 3 - 30t. On this side we're left with uh - 105 right and that will give the value for t5 over 30 that should give you 3.5. So the value of that capital T that they wanted to find wanted us to find that is 3.5. That's our final answer.

[01:22:44]Okay.

[01:22:51]Make sense, right?

[01:22:55]How do we know it's free fall? How do we know it's freef fall? Because the particle was projected vertically upwards.

[01:23:06]there is no resistance. If they don't tell us there is resistance, we assume there is no resistance.

[01:23:13]Okay? So weight is the only force that means it's free fall.

[01:23:19]Okay?

[01:23:24]Now they're saying find the height above O at which the particles collide. Now that's straightforward. We know the time at which the collision happens. Height above O, we can find either S A or SB.

[01:23:37]That will be the height at which the collision happens.

[01:23:43]Could say height is equal to displacement of A. We already have the function for that. This one a into T. t is 3.5 +/ into -10 into 3.5 squared and that will give you something. What's the value that we get from this? 218.75. Is that that an exact number?

[01:24:14]Yes. Okay. So, we can leave it as it is.

[01:24:16]Then if it's an exact number, 218.75 that's the height.

[01:24:22]Okay. All right. Let's move one now.

[01:24:26]Find the time from A being projected until C. What is C?

[01:24:34]Okay. They coales and become uh a combined particle C until C returns to O.

[01:24:43]A we already know that until the time of collision 3.5 seconds have passed. So that time we already know. So we are just concerned with what happens after a collision. We need to find that time.

[01:24:54]Right? Now how do we think about this?

[01:24:56]We've also found uh the displacement from ground at the point of collision.

[01:25:02]That's 218.75.

[01:25:05]First of all, let's see what happens at the point of collision.

[01:25:11]Particle A was we have to find the final velocities. Do we have them? No.

[01:25:18]Okay. So the first thing we should do is to use conservation of momentum. We should figure out the velocities of both the particles immediately before collision. Right? Immediately before collision we need to find their velocities. Let's find them VA and VB.

[01:25:39]What information do we have about both of them? For a we know that initial velocity was 80, acceleration was -10. We've also found displacement that is 218.75.

[01:25:58]And we also know time which is 3.5.

[01:26:04]So we've got all of this information. We want to find the final velocity. V = U + 80 will be really efficient.

[01:26:13]80 + - 10 into 3.5 and that gives you 45 m/s.

[01:26:22]That's the velocity at which A would be traveling just before collision. We can do the same thing for B as well. We know the initial velocity for A was 100.

[01:26:32]Displacement for this is also the same although we don't need that. uh acceleration is -10. What about time?

[01:26:40]It's time was t minus 1. That means 2.5.

[01:26:45]Find the final velocity for this. V = u + 80 100 + - 10 into 2.5. And that gives us 75 m/s. That's the velocity of B just before collision.

[01:27:00]Now B was projected later. So at the point of collision B will be below and A would be above. Right? So what do we know about the particles? Now we could say this is A. A was moving up at 45 m/s and B is at the same point as well at the point of collision and B is moving up at 75 m/s.

[01:27:29]Let's write down what information we've got about before and after.

[01:27:38]Before collision, let's consider one direction positive.

[01:27:43]We can consider upwards positive. Before collision and after collision for A and B. If you're considering upwards positive for A initial velocity is 45 for B it is 75 they're both going to be positive after collision they coales to form one single object they said their masses are yeah so the masses are equal their masses are equal so if this was m kg this is also m kg and what about the combined particle than ab mass of AB would be sum of those two numbers which is 2 m kg let's say its velocity is VAB in fact they called it C of the combined particle let's call it C this is VC apply conservation of momentum momentum before collision should equal momentum after collision what's the momentum before collision m into 45 + m into 75 that equals 2 m into bc m will get cancelled because it's appearing in every single term and you get the speed of the combined particle 120 /2 which is 60 m/s. That's the speed of the combined particle right after collision.

[01:29:14]Okay. Now what do we need to find out?

[01:29:15]We need to figure out the time until C. now hits the ground.

[01:29:23]So the situation is like this. After collision, let's suppose C is here.

[01:29:32]It is at a height of 218.75.

[01:29:36]We don't need to break it. We can do it directly. We did similar questions in the last couple of classes as well.

[01:29:44]218.75.

[01:29:46]If you want you can break it down uh that it reaches a maximum point uh at some point find that display find time taken for that and then from the maximum point to the bottom but it's not necessary for you to do that even directly you can say okay its initial velocity is 60 it's going in the upwards direction this is the starting point it goes like this and then comes back at some point and hits the ground now this is two-way motion In two-way motion, we could consider upwards positive. So, let's consider upwards positive for C for the combined particle. Now, let's consider upwards direction positive.

[01:30:28]What's the initial velocity? We know that is 60 m/s at which it starts moving in the upwards direction. Now, u is equal to 60. What about acceleration? We are considering upwards positive.

[01:30:40]So acceleration is going to be -10 because gravity acts in the downwards direction. Acceleration will be -10.

[01:30:48]We're looking for time. What else do we know? This is the starting point and this is the end point.

[01:30:57]Upwards is what we are considering positive. Displacement is in the downward direction.

[01:31:04]The starting point is above, end point is below. So the displacement value is going to be -218.75.

[01:31:16]That's the displacement value.

[01:31:20]All right. Now you can find the time taken.

[01:31:28]S= U +/ A². Let's use that equation.

[01:31:32]input values in this -218.75 that equals u which is 60 +/ into -10 into t ^ 2 now we get a quadratic equation 60 t - 5 t ^ 2 if I take all terms to the left side that will give me 5t ^2 - 6 dt - 218.75 = 0. Put this in the quadratic formula.

[01:32:08]You need to show working for that.

[01:32:10]Cannot write down the answer directly. - into -60 + -60 2 - 4 into 5 into -218.75 whole divided by 2 * 5. And that will give you two possible values of t. One is most likely going to be negative.

[01:32:35]Let's do that equation.

[01:32:38]You have those the answers. Okay. Thank you. 14.93 and -2.93.

[01:32:47]Now negative is of course not possible.

[01:32:49]So this is the value that we use. What's the total time taken now since a was projected?

[01:32:59]3.5 seconds had passed already before collision.

[01:33:03]So we can say now the total time taken is 14.93 + 3.5 and that gives us 18.43 or we can say 18.4 to3 significant figures. That's our final answer.

[01:33:40]make sense to everyone.

[01:33:42]Any questions on this?

[01:33:46]You see, we have seen similar questions in the last two papers as well or at least some at some point in the last week. So you should be able to do similar questions now in which you have to use constant acceleration formulas in one part initially and then collision happens you use conservation of momentum in that and then you have the combined particle moving again use equations of motion again. Yeah they're very similar but lengthy at the same time. You would only be able to do them in the exam if you do a lot of practice of such questions. Otherwise, they take a lot of time and time in M1 is pretty hard to manage.

[01:34:32]Okay. Question number seven. It says a particle P of mass 0.2 kg is projected vertically upwards from horizontal ground with speed 25 m/s.

[01:34:43]Show that the speed of P when it reaches 20 m above the ground is 15 m/s.

[01:34:55]Okay. So it's projected vertically upwards from horizontal ground. We can say upward direction is positive. We know the initial velocity. Initial velocity is 25.

[01:35:10]What else do we know? Upward velocity is positive. Right? Acceleration is going to be -10.

[01:35:19]when it reaches 20 m above the ground.

[01:35:21]That means when the displacement is 20, we need to show that the speed is 15. So, we have to find final velocity. V is what we're looking for.

[01:35:33]Now, which equation connects these four things? V ^2= U2 + 2 A S. And that gives us 25^ 2 + 2 into -10 into 20 and that will give the result. You can see what that is under root 225 which is 15. We get 15 from this.

[01:36:05]Yes. Okay.

[01:36:07]V equ= 15. That's the final answer.

[01:36:12]Yeah, of course. Yeah. So, we wanted to show that it's 15. That's the final result. So, now they're saying when P reaches 20 m above the ground, it collides with a second particle Q of mass 0.1 kg, which is moving downwards at 20 m/s. P is brought to instantaneous rest in the collision. Find the velocity of Q immediately after the collision.

[01:36:35]Okay. Now at that 20 m point, we now know the velocity of this first particle that was 15 m/s. So at the point of collision, the situation is like this.

[01:36:47]There's this particle P that is moving in the upwards direction at a speed of 15 m/s. And there's this particle Q that is moving at a speed of 20 m/s in the downward direction.

[01:37:11]We've got their masses as well. Mass of P is 0.2 kg. Mass of Q is 0.1 kg.

[01:37:18]This is 0.2 kg. And for Q it is 0.1 kg.

[01:37:25]We can use conservation of momentum.

[01:37:27]Take one direction. Take one direction as positive. Take one direction as positive.

[01:37:38]Which direction do we take positive? You can take upwards or downwards. Doesn't matter. Let's take upwards positive.

[01:37:43]That's more intuitive since we have particles moving in both directions.

[01:37:49]If I take upwards positive, I can say the velocities before and after. Let's just summarize them here first.

[01:37:58]For P and Q, what are the velocities before collision? For P, the velocity is 15. For Q, the velocity is -20. After collision, P is brought to instantaneous. This velocity is zero.

[01:38:11]What's the velocity of Q just after collision? Let's use conservation of momentum. Momentum before collision, that should equal Momentum after collision okay momentum before collision mass of P that is 0.2 into velocity of P before collision that is 15 plus mass of Q which is 0.2 into 0.1 sorry into velocity of Q which is -20 that equals momentum of P which is 0 after collision plus momentum of Q that's 0.1 * VQ and that is going to give it the value for VQ you can see what that is on the left side you get 1 or 10 1 okay 1 = 0.1 BQ and the value of BQ that is 10 m/s is that okay with Q is 10 velocity of Q just after collision that is 10 now it's positive that means it's moving in the upward direction okay after collision just after collision Q is moving in the upward Okay, make sense. Let's look at the last part now.

[01:39:46]Find when P reaches the ground, it rebounds back directly upwards with half of the speed that it had immediately before hitting the ground.

[01:39:57]Okay.

[01:40:01]When P reaches the ground, it rebounds back directly upwards with half of the speed that it had immediately before hitting the ground.

[01:40:11]Find the height above the ground at which P and Q next collide. Okay, that looks a bit tricky. Now, let's think about it.

[01:40:24]We had this horizontal ground. We had this horizontal ground. Now the situation is we know just after collision they are at a height of 20 m right from the ground. Let's just confirm that 20 m above the ground that was the height both the particles are at this height.

[01:40:45]After collision what happens is P starts moving downwards starting at a speed of 0 m/s.

[01:40:55]Q starts moving upwards starting at a speed of 10 m/s. We found that in the previous part.

[01:41:08]U is moving upwards at a speed of 10 m/s.

[01:41:14]Now what happens is P reaches the ground and then it rebounds back directly upwards with half the speed of what it had immediately before the ground. Let's talk about P first of all and figure out at what speed it goes and hits the ground.

[01:41:31]Okay, let's try to figure that out.

[01:41:35]For P we could say for P we could say the downward velocity uh downward direction is positive. What do we know about P? U is equal to zero.

[01:41:49]Displacement is 20. It's free fall right. We're taking downwards direction is positive. What's the value for acceleration going to be? Positive 10.

[01:41:58]So we can say acceleration is 10.

[01:42:09]Okay.

[01:42:15]Now we are looking for the final velocity.

[01:42:20]Let's find the final velocity. See what that turns out to be. V² equ= U2 + 2 S.

[01:42:27]Input values in this 0 2 + 2 into 10 into 20.

[01:42:34]And that gives us V = 20. This is the velocity at which P hits the ground.

[01:42:48]velocity at which B hits the ground.

[01:42:53]Now at what velocity does it bounce back?

[01:42:57]Half of that. Okay, it rebounds at half the half of that velocity.

[01:43:07]Okay, now we need to figure out the height above the ground at which P and Q next collide.

[01:43:23]Let's find the time taken for this as well because we also need to figure out how much has Q moved in the upwards direction while P was right P was direction while it was moving in the downward direction. Q would have also traveled some distance in the upward direction.

[01:43:41]Now how much has Q already traveled? We could find that.

[01:43:46]For that we need the time that P takes to hit the ground.

[01:43:50]Now, how do you find that time?

[01:43:53]V= U + 80. We could use that.

[01:43:58]V is 20. Initial velocity was 0.

[01:44:02]Acceleration was 10 into T.

[01:44:06]So, it takes 2 seconds for P to hit the ground.

[01:44:10]Q would have already traveled some distance in that time in the upwards direction. How much is that distance?

[01:44:15]Let's figure that out. I'm doing that on the right side here.

[01:44:20]Q was moving in the upwards direction.

[01:44:23]We can take upwards positive for that.

[01:44:26]Initial velocity for Q was 10.

[01:44:30]Acceleration would be -10. Since we're taking upwards positive until P hits the ground, the time taken is 2 seconds. Let's find the displacement that it has already traveled in that time.

[01:44:47]S = UT +/ A². We can say 10 into 2 +/ into - 10 into T ^ 2. You'll get some value from this. What's this value?

[01:45:03]What value do I get here?

[01:45:08]Zero.

[01:45:12]Okay. So what's happening is in those two seconds Q goes upwards and comes back to the same position in that time.

[01:45:23]All right. So that makes it slightly easier than from this point onwards.

[01:45:28]SQ we know that is going to be zero. If that is zero, can we also find the velocity at that point? If it's coming back to the same point, its velocity while going up was 10 m/s.

[01:45:41]We could directly write that the velocity while going down that will be negative 10 m/s. So basically 10 10 m/s in the downward direction.

[01:45:50]If you are not sure about that you can evaluate that as well. The velocity of Q at that particular point when 2 seconds have passed. Let's find that u + 80 u was 10 in plus acceleration which is -10 into time which is 2 and that will give you vq = -10. This is the velocity at which Q is moving when P hits the ground. So now we are at that stage at which P has already hit the ground and it is rebounding.

[01:46:31]and Q is moving in the downward direction.

[01:46:37]Okay, we are calculating time for Q.

[01:46:41]We're calculating time for Q because we need to figure out no we're not calculating time for Q yet.

[01:46:54]Uh time for Q was already given, right?

[01:46:56]We are calculating displacement and final velocity for Q.

[01:47:02]We are finding the displacement of Q until the point when P hits the ground. So P hits the ground after 2 seconds. In that time, we're figuring out where Q is at that point and what it what its velocity is at that point. Now that we figured it out, now it's a simpler problem and it looks like this. Q is here. I'm drawing that here again.

[01:47:34]After this point, now the situation is the particle Q is here and it is going to go downwards at a speed of 10 m/s.

[01:47:48]Then P was at the ground. P was at the ground.

[01:47:54]It rebounds [clears throat] at a speed that is half of the speed at which it hit the ground.

[01:48:01]That's what the question said. So P is rebounding at a speed of 10 m/s. So it's moving upwards in at the speed of 10 m/s. The displacement or the distance between the two particles that we've got at this point is 20 m. This is all that we know. Now we again apply those formulas. This is the situation.

[01:48:24]Now we could say from this point onward P and Q P is moving upwards. Now let's take upwards positive for P. Q is moving downwards. Now let's move downwards.

[01:48:36]Let's take downwards positive for Q.

[01:48:38]What information do we have about both of them? For P we know initial velocity is 10. Acceleration will be -10 because we're taking upwards positive.

[01:48:49]What else do we know?

[01:48:52]Nothing really. Let's uh write down the information that we've got about Q as well.

[01:48:59]What do we know about Q?

[01:49:01]About Q, we know initial velocity is 10. We're taking downwards positive for this. So acceleration will be positive 10 as well. These are two things that we know for both of them.

[01:49:17]Okay.

[01:49:18]Now about the other quantities, is there anything common from among the other quantities? time.

[01:49:30]They're starting at the same time instant in in a way. And until the point of collision, until the point of collision, both of them would have spent the same amount of time from this point onward. Let's say this is the point of collision.

[01:49:50]So time is common. If we say time for P is T, then the time for Q is also going to be T.

[01:49:57]And we also know if Q is coming down by a distance of SQ and P is going up by a distance of SP then the total of those two values is going to be 20. SP plus SQ that is going to equal 20.

[01:50:15]And that is going to give us the time value for this part of the journey.

[01:50:20]SP plus SQ that is equal to 20.

[01:50:25]Now what is SP?

[01:50:28]SP is UT +/ A² that that will be 10 T +/ into - 10 into T ^ 2 plus SQ that will be 10 into T +/ into 10 T ^ 2 that equals 20. Now these two terms get cancelled. They're both the same and we get 20 t equals to 20 and the value of t that we get from that is 1.

[01:50:59]So 2 seconds had already passed earlier then 1 second passes after that as well.

[01:51:08]What we want to do is we want to find the height above the ground. Okay. Once we've got the value of t.

[01:51:14]I thought we were looking for time. No, we have found the value for t. Now from this point onward we can find the displacement of p. Now the height above the ground at which this collision happens this height this is the same as the displacement of P. So we can find the displacement of P.

[01:51:30]Now from the ground for P we know initial velocity was 10 acceleration was 10. We al also have the time value. Now let's find the value of T. UT +/ A² that will be 10 into 1 +/ into - 10 into T² that will be 1 2 and that will give you the result which is 10 - 5 and that is 5 m that's the height above the ground at which that next collision happens. That's what you were looking for.