Question

*a small object with a momentum of 6 kg∙m/s to the west approaches head-on a large object at rest. the small object bounces straight back with a momentum of 5 kg∙m/s. what is the change in the momentum of the small object? what is the impulse exerted on the small ball? what is the impulse exerted on the large object?

Answer 1

1) The change in the momentum of the small object is given by the difference between its final momentum and its initial momentum:
$\Delta p = p_f - p_i$
where $p_f, p_i$ are the final and initial momentum of the small object. Substituting the numbers of the problem, we find
$\Delta p = - 5kg m/s -(6 kg m/s) = -11 kg m/s$
where we put a negative sign in front of pf because the direction of the object after the collision is opposed to its initial direction.

2) The impulse is equal to the product between the force applied to an object and the time interval:
$J= F \Delta t$ (1)
since the force is the product between the mass and the acceration: F=ma, and the acceleration is equal to the change in velocity divided by the time, we can rewrite (1) as 
$J=F \Delta t=ma\Delta t=m \frac{\Delta v}{\Delta t} \Delta t = m \Delta v = \Delta p$
So, the impulse is equal to the variation of momentum. Therefore, the impulse exerted on the small object is
$J=\Delta p = -11 kg m/s$

3) We have to find the final momentum of the large object. Initially, its momentum is zero, because it is at rest. Using the law of conservation of momentum,
$p_i + 0 = p_f + P_f$
where $p_f, p_i$ are the final and initial momentum of the small object, while $P_f$ is the final momentum of the large object. By substituting the numbers we had before:
$P_f = p_i - p_f = 6 kg m/s -(-5 kg m/s) = 11 kg m/s$
So, the impulse exerted on the large object is equal to its variation of momentum:
$J= \Delta P = P_f - P_i = 11 kg m/s -0 = 11 kg m/s$