A cart of mass 0.65 kg initially moves right at 1.5 m s⁻¹. A sensor records the net horizontal force, with right positive. The force-time graph is straight between these (time in s, force in N) points: (0, 0), (0.15, -16.25), (0.3, 0). After the final point the force is zero. Find the impulse, final velocity and change in kinetic energy. State whether the cart reverses direction and whether the interaction must remove kinetic energy.

Question

Accepted Answer

Impulse -2.438 N s; final velocity -2.25 m s⁻¹; ΔK = +0.9141 J. It reverses direction and gains kinetic energy. Impulse is the signed area under the net force-time graph. Here the pulse lies below the axis, so its impulse points left even though the cart initially travels right. The triangular pulse gives J = ½F_peak Δt = -2.438 N s. The initial momentum is mu = 0.975 kg m s⁻¹. Apply J = mv − mu. Thus v = u + J/m = -2.25 m s⁻¹. The negative sign means that the cart has reversed and now travels left. Kinetic energy depends on speed squared, not the sign of velocity. ΔK = ½m(v² − u²) = +0.9141 J, so kinetic energy increases overall. The cart first slows, stops and then accelerates left while the pulse continues. A negative impulse therefore does not imply negative net work; the agent producing the force can supply energy after the reversal.

A force pulse reverses a cart · set 5

What a complete response should do

Recorded net force on the cart

How did this one feel?

Recorded net force on the cart — supplied values
Time / s	Net horizontal force / N
0	0
0.15	-16.25
0.3	0