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…
$0.4(3) + 0.8(-100) = 0.4(-2) + 0.8(v)$
$v=-97.5m/s$
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.
Best Answer
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.