Collision and Impulse — Question 7

Question 7

A force sensor measures one member of a pair

During contact between carts A and B, the force on A is $F_A(t)=120\sin(\pi t/T)\,\mathrm N$ for $0<t<T=0.050\,\mathrm s$. The carts have masses $2$ and $3\,\mathrm{kg}$. External horizontal impulses are negligible.

1. Find the impulse on each cart and each velocity change.

2. Compare peak forces and peak accelerations. Which equality follows from Newton's third law?

Question 7 - Solution

A force sensor measures one member of a pair

(a) Derivation and calculation

Integrate the sine pulse and use the equal opposite contact force: $$J_A=120\int_0^T\sin(\pi t/T)\,dt=\frac{240T}{\pi} =\boxed{\frac{12}{\pi}\ \mathrm{N\,s}},\qquad J_B=-J_A.$$ $$\boxed{\Delta v_A=J_A/2=6/\pi\ \mathrm{m/s}},\qquad \boxed{\Delta v_B=J_B/3=-4/\pi\ \mathrm{m/s}}.$$

(b) Result and interpretation

At the pulse maximum, $$\boxed{|F_A|_{\max}=|F_B|_{\max}=120\ \mathrm N},\qquad \boxed{|a_A|_{\max}=60,\quad |a_B|_{\max}=40\ \mathrm{m/s^2}}.$$ Newton’s third law equates opposite forces and impulses, not the accelerations of unequal masses.

Physical check

The weighted velocity changes cancel: $2(6/\pi)+3(-4/\pi)=0$.