Collision and Impulse — Question 9

Question 9

Equal impulses delivered at different times

A $1\,\mathrm{kg}$ cart starts at rest. In experiment A a $4\,\mathrm N$ force acts over $0<t<1\,\mathrm s$; in B it acts over $1<t<2\,\mathrm s$. Otherwise horizontal force is zero.

1. Compare final velocities at $t=2\,\mathrm s$.

2. Compare displacements by $t=2\,\mathrm s$ and explain why the impulse theorem alone cannot determine them.

Question 9 - Solution

Equal impulses delivered at different times

(a) Derivation and calculation

Both experiments apply the same total impulse by two seconds: $$J_A=4(1)=4\ \mathrm{N\,s},\qquad J_B=4(1)=4\ \mathrm{N\,s}.$$ $$\boxed{v_A(2)=v_B(2)=0+J/m=4\ \mathrm{m/s}}.$$

(b) Result and interpretation

The earlier pulse gives an extra second of coasting: $$\boxed{x_A(2)=\tfrac12(4)(1)^2+4(1)=6\ \mathrm m},\qquad \boxed{x_B(2)=0+\tfrac12(4)(1)^2=2\ \mathrm m}.$$ Generally, from rest, $x(T)=m^{-1}\int_0^T(T-t)F(t)\,dt$; displacement retains the timing that total impulse discards.

Physical check

The final velocities agree while the earlier acceleration produces 4 m more travel.