Question

A $0.2kg$plastic cart and a $20kg$lead cart can both roll without friction on a horizontal surface. Equal forces are used to push both carts forward for a time of $1s$, starting from rest. After the force is removed at $t=1s$, is the momentum of the plastic cart greater than, less than, or equal to the momentum of the lead cart? Explain.

Short answer

equal

Step-by-step solution

Given information

We have given that

We need to find

Simplify

Newton's law is formulated in terms of momentum, where the forces is the change of momentum with time. Where a force is exerted on an object, it changes its momentum and the force is given by equation $(11.4)$in the form

$F_x=\frac{d{p}_x}{dt}$ $\left(1\right)$The plastic and lead carts are differ in masses, but we apply the same force $F_x$on both of them , so equation $\left(1\right)$for the two carts will be

$$\begin{gathered} F_{plastic}=F_{lead} \\ \frac{d{p}_{plastic}}{dt}=\frac{d{p}_{lead}}{dt} \end{gathered}$$ $\left(2\right)$

Also, the force is delivered to both carts in time $t=1s$.And as time is the same, the momentum as shown by equation $\left(2\right)$will be the same

$$\begin{gathered} \frac{d{p}_{plastic}}{1s}=\frac{d{p}_{lead}}{1s} \\ {P}_{plastic}=P_{lead} \end{gathered}$$

Hence, the momentum of the plastic cart equal to the momentum of the lead cart.