Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 11 - Impulse and Momentum - Exercises and Problems - Page 286: 4

Answer

The momentum of both the plastic and lead carts are equal.

Work Step by Step

Newton's second law states that the rate of change of the momentum vector gives the force vector acting on a body. Mathematically, $\vec{F} =\frac{\mathrm{d}\vec{p}}{\mathrm{d}t} $. To find the change in momentum due to the application of a force $\vec{F}$ from a time $t=0$ to $t=t_f$, we need to integrate the above differential equation to obtain the impulse defined as $\vec{J} = \int_0^{t_f}\mathrm{d}t \vec{F} = \Delta \vec{p}$. In the above case, an equal force(it may or may not be constant) is used to push both the carts from $t=0$ to $t=1\mathrm{s}$. Since the forces are equal and the time of application is the same in each case, the impulse of the force, and thus the change in momentum of the carts, will be the same in both cases. Both carts started from rest and thus, will have the same final momentum after the removal of the force. An important point to note here is that since the plastic cart has a much smaller mass compared to the lead cart, it will have a higher velocity than the lead cart. However, the momentum $\vec{p} = m\vec{v}$ will be the same in both cases.
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