Master 9231 Further Mechanics Momentum FAST | Complete Syllabus Revision

Added: 2026-05-20

[00:00:00]Everybody uh welcome to my channel.

[00:00:02]Today we're going to be learning momentum from further maths. So let's get started. So first of all you have to understand what a momentum actually is.

[00:00:14]It is basically the product of mass and velocity. Okay.

[00:00:19]So so what could a momentum actually mean right? A momentum is basically linked to its force. Suppose a bowling ball moving with a very slow velocity. Let's say 1 m/s uh hitting a wall could have the same impact as a tennis ball moving at a very fast velocity of 5 m/s hitting a wall. It could exert the same amount of force meaning the pressure exerted. Okay, it basically means that a mass is interlin to its force. Meaning the higher the mass we have and the higher the velocity we have the more force we will exert. Okay. So thinking in terms of that momentum can uh force can be defined as the rate of change of momentum. Okay.

[00:01:25]So what is an impulse? Impulse basically means the change in momentum in an object. Okay. So in short impulse is the product of force and time or the change in momentum. All right.

[00:01:43]This formula is defined just like this.

[00:01:45]We can say that uh force is equals to mv minus mu which basically means the change in momentum divided by t. So mv - mu = to f_t.

[00:02:03]We already know that the change in momentum is just the impulse. So I can be written as the impulse. Okay.

[00:02:13]So talking about direction uh let's say a ball is moving towards the wall and uh coming outwards the wall. Okay. So on the ball by the wall the impulse is going to be mv minus mu obviously because it is just the change in momentum. But when it's on the ball by the wall basically the wall is hitting the ball. So the left side is actually positive.

[00:02:49]In that sense our mv should be positive and our mu should be negative. So we can see that it's already just like that. So it should give me a value of positive.

[00:03:02]Okay. But when it's on the wall by the ball meaning it's moving towards the right in that sense we can say that our u is positive and our v is negative right so we can see that uh this should give me something like this which will end up giving me a negative impulse.

[00:03:28]Okay. So that's the thing about direction we might have to worry about.

[00:03:34]Now conservation of momentum. We have already learned in M1 that the total momentum before is actually equals to the total momentum after. Okay. So writing that in terms of formula we can say that m_sub_1 u1 + m_sub_2 u2 is actually equals to m_sub_1 v1 plus m_sub_2 v2. Okay.

[00:04:00]So now we kind of starts to move towards the further maths concept of momentum.

[00:04:08]All right.

[00:04:10]So Newton's experimental law tells us that the speed of separation divided by the speed of approach gives us equation I mean coefficient of restitution. Okay.

[00:04:28]So what the coefficient of restitution actually tells us is how the objects will collide. So let's say that our E is zero right if our E is zero and our object collides our object will form together co and we can say that it colles okay but if our E is a value of one we can say that our object perfectly collides all Right?

[00:05:06]It perfectly collides and yeah that's it. Basic as I was saying if the E is 0.5 we can say that some of our velocity of the object will be kind of lost. Let's say it turns down to half a velocity. I hope that makes sense. when it's actually E, our speed actually doesn't changes at all.

[00:05:37]But when E is actually equals to zero, we can say that it it is perfectly in inelastic, which mean it basically collaces. Okay. So it that basically means that our maximum kinetic energy is lost. But when it's perfectly elastic, no kinetic energy is lost. But when it's somewhat in between uh some kinetic energy will be lost. Okay.

[00:06:08]So now let's learn oblitions.

[00:06:12]So for oblitions it basically means an object colliding with a while. Let's say an object collides with a wall with a speed of v.

[00:06:24]Okay. So if that object Oh, I'm sorry.

[00:06:30]So if that object comes back with a speed of u that basically means that the object comes back with a speed of eu.

[00:06:43]Okay, let me explain that again.

[00:06:49]Let's say an object moves with a speed of u. Okay. And if the object comes back by bouncing with the wall, the speed of our object will be something like V and that V will be the product of equation of restitution into U. Okay. So in this case our wall is vertical, right?

[00:07:19]So we can see that the speed at which the speed at which my object would collide should correspond with the horizontal velocity. Okay. So you can see that our original horizontal velocity is u sin theta which I'm just going to consider as ux. So we can see that it collides here and it comes back just like this. We know that if it collides with a while just like this, it comes back with a speed of e multiplied to the initial speed. So we can say that our horizontal component here will just be e of our initial speed and our vertical component will just say stay the same.

[00:08:17]Now in this case we can see that it's actually uh the while is horizontal.

[00:08:24]So if an object collides with it, it obviously corresponds to it vertically, right?

[00:08:32]So our initial vertical component is just uy which it hits and then it collides and goes up with a speed of e of our initial velocity.

[00:08:50]I hope bouncing with a wall makes sense.

[00:08:53]Now in terms of oblique collisions.

[00:08:56]All right. So now we have angle of deflections.

[00:09:01]So let's say an object collides with another object with an angle.

[00:09:10]Right? This is the angle we're talking about.

[00:09:16]So A was initially moving like this and B was just at rest. All right. So we can say that when it collides the horizontal component works towards this and the vertical component just stays like this. Right?

[00:09:36]Okay. So this is just the collision before. Okay. So after the collision what happens is the block that does collide it angle changes. So it was initially theta it now becomes alpha. Okay.

[00:10:01]So uh I should I think I made a mistake here.

[00:10:15]This should be u sin theta.

[00:10:24]Yep. All right. So now we can see that our horizontal component has hit the particle here.

[00:10:34]So what should happen is the my particle B should moves towards right since there was no vertical component acted to this object.

[00:10:54]It should not have any vertical components. Okay.

[00:11:00]But for our A, it should obviously lose some of its speed since it transferred some of it to our object B.

[00:11:13]So its horizontal component actually changes with respect to its angle. Okay? and its vertical component just stays the same since no vertical collision acted on it. Okay.

[00:11:33]So what we can say here is our this alpha if we just make a simple triangle we can see that is u sin theta which is just the v y and this value is v cos alpha. In terms of that I can just say that tan inverse is vy over vx.

[00:12:04]Okay. And for the angle of deflection let's say the a hit like this. And now it's traveling towards this right. So for the angle of deflection we can just say it's for theta.

[00:12:23]Okay. One thing I missed was uh the speed of a after the collision can be written as uh u y square plus vx² whole square of root. Uh this is basically just the pythagoras theorem.

[00:12:44]We already know that the uy is uh u sin theta and vx is v cos the uh alpha. Okay. So the speed of a can be found find found out just like that.

[00:13:02]Okay.

[00:13:03]I hope uh the formulas at least made a little sense. If not it's fine. uh we are going to go through all the example maths and it should make sense at the end of the video.

[00:13:23]So in example 18.1 a particle of mass 4 kg mass 4 kg is traveling in a straight line on a smooth horizontal surface. It basically means that a particle is traveling on a smooth horizontal surface. The particle is moving at constant velocity of 3 m/s when a force of 5 Newton is applied for 0.4 seconds. Find the impulse applied to the particle and its velocity after 4 seconds. The particle does not change direction during its motion. All right.

[00:14:17]So first of all, we can just directly use the formula of impulse of force into time and find out the impulse by multiplying this and this. Okay, so that was super easy. So we have to actually understand the concept. So our particle is moving in a horizontal surface, right? So its initial velocity is just u equals to 3.

[00:14:53]So after it has traveled 0.4 seconds, they asked me to find its velocity.

[00:15:03]Right? So we know that impulse is basically the change of momentum. So we know that impulse is two.

[00:15:13]The final velocity we don't know.

[00:15:17]And the initial velocity we do know as three. So 4 V is just equals to 14 and V should be 3.5 m/ second. Okay.

[00:15:37]All right.

[00:15:40]So for the second question, a particle of mass 10 kg travels in a straight line. Again the particle travels in a straight line. The particle is traveling with a velocity of this. A force of magnitude is applied to the particle for 12 seconds. Given that the particle direction is reversed, find the velocity after 12 seconds. So what happens here is a particle was moving with a velocity of 12 m/s but now some guy pushed the particle with a force of 18 Nton right so after that our particles the direction is just reversed so our particle moves in a velocity towards wards this direction right?

[00:16:40]So what we can say is during this 12 seconds what is our velocity after right so of course we know the force we know the time we're going to find the impulse using force into time and then we have to find the final momentum so Since our direction is the opposite, what we can do is we're going to take the left side to be positive and the right side to be negative.

[00:17:26]So this initial velocity is of 12.

[00:17:33]And we know that change in momentum is just mv minus mu. And we did our algebra to find the velocity of 9.6.

[00:17:42]I hope that makes sense.

[00:17:45]Okay. A particle of mass 5 kg.

[00:17:49]Okay. A particle of m mass 5 kg is traveling along a smooth horizontal surface with velocity of 15 m/s.

[00:17:58]Given that the force F applied over 4 second reduces the velocity by 8 m/s.

[00:18:06]Okay. And the particle still travels in the same direction.

[00:18:12]Find the force of F.

[00:18:14]Okay. So, a particle had a velocity of 15 m/s.

[00:18:23]It goes to the same direction, but it velocity decreases to 8.

[00:18:32]So obviously a force was applied to the opposite direction right. So what we can say is our force is negative since I have taken the right side to be positive. Okay.

[00:18:54]So after taking the force to be negative and we know the time is just force I mean the time is four and using mv minus mu taking the correct direction we can find the force to be 10 nton. I hope that makes sense.

[00:19:14]Okay.

[00:19:15]Now all of these were just the type one for impulse.

[00:19:29]Very very easy.

[00:19:37]Second type of maths. Uh this is basically the type two Newton's experimental law. This is very very common in your cushion paper. So let's get started. So here you can see that the collision is perfectly elastic.

[00:19:55]We have initial velocity of 3 u and u and we don't really know the final velocity for both the particles.

[00:20:06]So we do have to find the velocity first. So yeah let's do that.

[00:20:11]So first of all we can do conservation of momentum.

[00:20:17]By doing conservation of momentum we found this equation.

[00:20:22]This was really simple to find out. And after that we can see that the uh collision is perfectly elastic.

[00:20:31]So when the collision is perfectly elastic we can say that e is basically equals to 1.

[00:20:40]So after that we can see that the speed of approach is basically 3 u minus u. Why? Let's go back to the formula. We can see that the speed of approach is v_sub_1 minus v_sub_2 and the speed of separation is v_sub_2 minus v_sub_1. Right? So you can see that I wrote the speed of separation as v_sub_2 minus v_sub_1. But here uh for speed of approach it's actually v_ub_1 minus v2 where v_sub_1 is basically 3 u and v2 is basically u.

[00:21:22]So there we go. I put that value in.

[00:21:26]After that I found another equation.

[00:21:30]So we can see that the u here is common and we have to find v_sub_2 and v_sub_1.

[00:21:37]So after doing simultaneous equation I found v_sub_1 and v_sub_2 in terms of u.

[00:21:44]All right. So after that they ask me to confirm that the kinetic energy is the same before and after the collision.

[00:21:53]So before the before the kinetic energy is basically kinetic energy of this and kinetic energy of this which is basically half mv²/ half mv². I simply wrote that and I just added it up. Right? So I got 14 U².

[00:22:15]So after I found V_sub_1 to be 2 U and V2 to be 4 U. So basically after I have to take this kinetic energy and this kinetic energy. So after adding them up again with our different speed I can see that this is also 14 new square. So I can finally state that they are the same.

[00:22:43]I hope that makes sense to you guys and this question so should be done. So in this question um they have given me two spears on a smooth horizontal surface.

[00:22:56]Okay.

[00:23:01]Okay. And they have told me that the equation of restitution is 2x3 which is somewhat around 0.667.

[00:23:12]So with this knowledge I can definitely say that there is some loss in velocity.

[00:23:19]So definitely kinetic energy must be lost as well. Right? So first of all they wanted me to find the velocity. So remember when we do not know the final velocity of both the objects and we have the equation of restitution.

[00:23:40]First of all, we have to use princip principle conservation of momentum to make an equation.

[00:23:49]So I just used this formula to make this equation using the directions I have. So here this will be negative since I took right as positive. So I found this equation I mean equation one and then I already know that the equation of restitution is 2x3 which means when the object collides it collides with an E value of 2x3 that means velocity must be lost. So E is basically the speed of separation divided by the speed of approach. Here speed of separation if we go back to the formula we can see that speed of separation is v_sub_2 minus v_sub_1 so I wrote v_sub_2 minus v_sub_1 and the speed of approach is v_sub_1 minus v_ub_2 so here you can see that v_sub_1 is basically 2 u right and u is in the opposite direction right I took right as positive so this must be negative of u which is basically 2 u + u and since we got this so after this we got this equation here I can see that u is common and I have to find v_sub_1 and v_sub_2 so I just use simultaneous equation to find both of this answer and after that I found the kinetic energy before with the initial speed I had for both the particles which I found to be 11 by two and with the new speed I have found I found the kinetic energy after and which I did find to be 5 / 2 and they asked me to find the loss in kinetic energy so it should be 11 by 2 - 5x 2 this should be the kinetic energy that has been lost all right so in this question the value of E is actually not given but um the initial velocity of P has been given by U as you can see and it has been clearly written that my object Q is resting. So this must be 0 m/s. Okay.

[00:26:18]So after the collision the E has not been given. All right. So since P was moving towards here, I have to find a value I mean find the values of E for which the P will keep moving towards the right. All right. So for what I did here is first of all I found an equation using principle of conservation of momentum. That's my equation one. And then I use the same concept to find uh this equation in terms of E. All right.

[00:27:03]Here you can see that I took V2 plus V1.

[00:27:07]Right? Why? Because I took the V1 to be moving towards the left. Okay?

[00:27:18]So why I took the V1 towards the left is because it asked me that the direction of P must be changed. All right. So I can just write that V2 minus negative of V_sub_1 which just gives me this. All right. So after that I use simultaneous equation to find both of these values.

[00:27:45]So this v_ub_1 must be greater than zero because if it is negative if it's negative it just moves towards the right. So I took this v1 and I took it to be greater than zero. After that I used algebra to find e greater than 1 / 2.

[00:28:11]So for the condition where equation I mean coefficient of restitution is greater than 1 /2 the p must change its direction to the left. If it's less than 1 /2 it will just keep moving towards the right. I hope that makes sense. Okay. In this question a particle has been projected with a velocity u. So here a particle has been projected right. So it has also been given that the angle with the collision of the W is 60°. So if this is 60° this must be 60° as well due to alternate angles right. So after it collides it has been given that it must have an angle right. So I took the angle as alpha you can take theta or whatever.

[00:29:08]So if this is alpha this must be alpha as well due to corresponding angle right. So here the horizontal component is u sin 60 since this is the initial speed and this is the angle.

[00:29:25]So if it collides like this and it goes back like this this must be e of initial component right? So, yep, this is just my E of my initial component.

[00:29:42]So, what they have asked me to find is the angle, right? So, how are we supposed to find the angle? So, for this E value, they have already given me that it's 0.3.

[00:29:56]So, what I have did here is I have found the horizontal component first and I have also found the vertical component.

[00:30:04]Right? So we already know that um this tan alpha is basically if you make a separate triangle this tan alpha is just vx over vy opposite by adjacent. Right? So I wrote that we know that this vx this vx is basically e of our initial horizontal component right so I wrote that as well and we also know that our vertical component never changes.

[00:30:48]So I can also write that this is basically u cos 60 and u cos 60 is half of u and u sin 60 is this. So after putting my values u and u basically gets cuts off and I just used tan inverse and I found the angle to be 27.5°.

[00:31:13]Okay, in this question um the particle here is traveling on a smooth horizontal surface, right? But then it is given that it oblitely collides with a smooth vertical wall. Here's the vertical wall.

[00:31:29]So here's the equation I mean coefficient of restitution is given by by 1 / 3. So I can say that this should be e of ux.

[00:31:41]So they have told me that the angle with the wall is 40°.

[00:31:46]So what I did here is resolve these right I have already said why these angles are same. So what I have did is resolve all of these and we do not know the final velocity right. So I didn't take the final velocity form now. So first of all what I did here was I found the initial kinetic energy.

[00:32:14]You can use this formula as well. Basically roo<unk> over of ux + u y² but it will just give end up giving you four since this and this is basically one.

[00:32:32]So you don't have to do that. You can just take the speed directly and find the kinetic energy before. So for to find the speed after I can say that my speed should be using Pythagoras theorem I can say that my speed must be this is vy this is vx using pythagoras theorem I can say that this is vy whole square + vx whole square right we know that vy does not change with uy so I took uy here and we know that The horizontal component is basically the product of E and the initial horizontal component.

[00:33:15]Right? So after that I have found the velocity to be this. So after that I have found the kinetic energy after using this speed. Okay. So I can see that my kinetic energy compared to this has lost quite a bit. So I just wrote the loss of kinetic energy with the difference and that's my answer. Okay.

[00:33:41]Okay. So in this question uh they have already said that it obligely collides with a smooth vertical wall. So it also rebounds with an angle of 90°. So if it rebounds like this right this angle must be 90 and they have said that the initial angle with the while is 60° so this is 60 you have to make sure that this velocity is moving like this I just used this like this method here this scenario you can take the opposite one as well it's no problem so if this is 60 and this is 90 This angle must be 30° since this is a straight line. So what I have did here is actually this should be like this vx.

[00:34:42]Okay.

[00:34:46]>> So um my final velocity must be the product of e and the initial velocity. Right? Since it collides like this and comes back like this. So my final velocity I can write as V sin 30.

[00:35:03]I just resolved it. And we do not know the E value of E. We do know that the initial component is this. Right? So after that I can say that uh vy and uy are basically the same which is just u cos 60°.

[00:35:23]So finally I know that this angle is 30° and if I make a triangle this being 30 uh this is my vx and this is my vy I can say that this is vx by vy right so finally after that uu gets cuts out and I can just multiply this with this and find the value of e. All right, I hope that makes kind of sense. Okay. Okay. So for the final type of question in this question, um my object P collides with my object Q. So let's read the f uh question first.

[00:36:29]All right. So you have to visualize uh the question kind of like this. So my object P collides with object Q at an angle of 60° right but only my horizontal component will act on object Q. Okay. So this was initially at 60° right? So my horizontal component acts here. So this speed will obviously uh be changed and be decreased.

[00:37:04]So this if this was ux this will now be vx and this I will be taking as a v2. Okay. And my uh vertical component will just stay the same. I don't know this velocity and I don't know this angle as well. And in the question they have asked me to find the angle of deflection. Okay. So first of all what I did was use principle of conservation of momentum here with just the horizontal components acting. All right with just the horizontal component acting. I know that the equ uh coefficient of restitution is 1x4. So I took speed of separation and speed of approach and I found another equation here. So after that I use simultaneous equation to find both the velocities.

[00:38:02]Right? So here I only need V1 since I have to find the angle of deflection. So I took the v_sub_1 here. So if you make a triangle now this is v_sub_1 and this is vy. I can say that tan alpha is v y over v 1 right so that's exactly what I have written here and we already know that vi is just the uh initial uh vertical component so I took both of these and I found the angle to be 87.9° and if we go back to the formula my angle of deflection is alpha minus theta, right? And my theta was obviously 60°. So my angle of reflection must be 87.9 - 60, which is just 29 27.9°.

[00:38:59]And that should be everything about momentum. In the next video, I will do past papers from 2024 to 2021.

[00:39:12]So, make sure to wait for that video and watch it as well. Okay, so that should be it. Thank you for watching. Goodbye.

[00:39:21]Uh, leave a like. Uh, leave a subscribe as well. Uh, that should help the algorithm. And yeah, thanks for watching. I hope it really helps