Collision and Impulse — Question 3

Question 3

Signed pulse areas can cancel

A $2\,\mathrm{kg}$ cart initially moves at $1\,\mathrm{m/s}$. Its net horizontal force is $+6\,\mathrm N$ for $0<t<1\,\mathrm s$ and $-3\,\mathrm N$ for $1<t<3\,\mathrm s$.

1. Find the impulses of the two stages and the final velocity.

2. Find the largest kinetic energy reached. Does zero net impulse imply no kinetic-energy change during the process?

Question 3 - Solution

Signed pulse areas can cancel

(a) Derivation and calculation

Compute both signed impulses: $$J_1=6(1)=6,\qquad J_2=-3(2)=-6\quad\mathrm{N\,s}.$$ $$v_1=1+J_1/2=4\ \mathrm{m/s},\qquad \boxed{v_f=1+(J_1+J_2)/2=1\ \mathrm{m/s}}.$$

(b) Result and interpretation

Velocity rises during the first stage and falls during the second, remaining positive: $$\boxed{K_{\max}=\tfrac12(2)(4)^2=16\ \mathrm J}\quad\text{at }t=1\ \mathrm s.$$ $$K_i=K_f=\tfrac12(2)(1)^2=1\ \mathrm J.$$ Zero net impulse restores final momentum but does not prevent intermediate energy changes.

Physical check

The first stage adds 15 J and the second removes it.