I was running over a question,

A collision occurs between particles A and B which are moving in opposite directions in the same straight line. The impulse on each particle in the collision is 2N.s, it is also given that A has 0.4 kg, and initial velocity 3 m/s. B's velocity changes by 2.5m/s. So what it wants us to do is find the final velocity of A and mass of B, that's easy enough. Lets say anything going right is positive and A is moving to the right.

To find final velocity of A:

  • we simply use the impulse, I = m(change in v), which gives us -2 m/s.

To find mass of B:

  • do the same thing...and we get 0.8kg

This is where I get confused, the last part says deduce a maximum of initial speed of B. Technically if the impulse of both particles are the same, then it doesn't matter what the initial speed of object B is, since impulse is still 2N.s.

For example:

if B has initial velocity 100 m/s, and keeping all else the same, and using I = m(change in v)

$2 = 0.8(-v-(-100))$

$v = -97.5m/s$

If we use conservation of momentum, same thing...

$0.4(3) + 0.8(-100) = 0.4(-2) + 0.8(v)$


Obviously in real life, we would most likely see a larger change in the speed of A and B. However, the answer is 4.5m/s, and I still fail to see why there will be a maximum speed if the impulse is the same for both objects.

  • $\begingroup$ is this a elastic or inelastic collision? $\endgroup$ – Jaywalker May 27 '16 at 11:31
  • $\begingroup$ Ok I just did some calculations and this has to be an inelastic collision because KE is not conserved. That may be the key to the answer ;) $\endgroup$ – Jaywalker May 27 '16 at 11:41
  • $\begingroup$ yes, but that's not the point. this part of the book comes before them using KE, that's why decided not to use it. $\endgroup$ – user51515 May 27 '16 at 17:45
  • $\begingroup$ even if i decided to use KE, i don't see how that would matter, since B had a higher starting initial velocity. And if KE is not conserved regardless of initial starting velocity, it wouldn't apply to the previous case either when B had a starting lower velocity. So that doesn't answer the question. $\endgroup$ – user51515 May 27 '16 at 17:57
  • $\begingroup$ nvm, i figured it out. how stupid of me. if the speed of A is 2m/s after the collision, B cannot be travelling faster than that if another collision does not occur, therefore the speed is 4.5m/s before. Thanks for your help, but would appreciate it a bit more if you showed the calculations, and not say things like "it has to do with KE", you need to explain why, in any case, it has nothing to do with energy. How were your IB exams? $\endgroup$ – user51515 May 27 '16 at 18:35

This is a tricky question. I think you are all right and this problem, because using momentum, is valid for both elastic and inelastic problem.

the tricky part is, particle A cannot penetrate particle B. If, they collide, particle A moves at -97.5m/s and particle B moves with -2m/s, do you think that is possible?

So if particle A initial velocity is above a maximum speed 4.5m/s, the impulse cannot be 2Ns.

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Let's say we have a collision between a moving object A and a stationary object B, of which A has a mass of 25kg and B a mass of 50kg.

The collision is perfectly central and 'elastic' e.g. with no additional internal absorption or energy conversion. [ the impulse exchange is pure ] meaning there is absolutely no transitive physical deformation between the colliding objects, neither elastic, nor plastic during the collision.

What will be the resulting inertial state of the system for objects A and B after the collision if the speed of object A was, say 28 m/s?

mA = 25kg; vA = 28 m/s 
mB = 50kg; vB =  0 m/s

After a perfect central collision course, we'll have:

vA = ? 
vB = ? 

It is obvious that object A will get reflected i.e. deflected from its initial course, and continue to move in the opposite orientation of its original direction, but will manage to deliver a certain amount of its impulse to the resting object B which will assume its moving orientation.

What will it be?

Let us first examine a case of the same scenario by merely switching the masses. In this contrasting case, the object A has twice the mass of the encountered resting object B.

This is a bit easier. (Of course, now the total amount of P (impulse) of the system is twice as large for the same speed of the object moving into a collision course ).

In this case the twice as massive object A, will hit the resting object B. No deflection will occur. That is, the object B, will immediately assume the moving orientation of a two times heavier object A. Object A will preserve some of it's moving direction and orientation of the remaining impulse.

What will be the resulting inertial state (speed) of object A, after colliding with object B?

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I found that if the initial speed of B (in the negative direction) was too large, then the total energy after the collision was greater than before.

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