Question

question 5 (4 points)
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a 5 gram bullet moving 328 m/s is shot into a 2.01 kg block, which slides for 1.54 seconds across the surface it is on before coming to rest. what is the average force of kinetic friction between the block and the surface?
your answer:
answer units

Explanation:

Step1: Apply conservation of momentum

The initial momentum of the bullet is \(p_{bullet}=m_{bullet}v_{bullet}\), where \(m_{bullet}=5\ g = 0.005\ kg\) and \(v_{bullet}=328\ m/s\). The block is initially at rest (\(v_{block}=0\)). After the collision, the combined mass is \(M = m_{bullet}+m_{block}=0.005 + 2.01=2.015\ kg\). By conservation of momentum \(m_{bullet}v_{bullet}=(m_{bullet}+m_{block})v\), so \(v=\frac{m_{bullet}v_{bullet}}{m_{bullet}+m_{block}}=\frac{0.005\times328}{2.015}\approx0.816\ m/s\).

Step2: Use impulse - momentum theorem

The impulse \(J = F_{avg}\Delta t\) is equal to the change in momentum \(\Delta p\). The final momentum is \(0\) (comes to rest), and the initial momentum of the combined system is \(p = Mv=(2.015)(0.816)\approx1.644\ kg\cdot m/s\). Since \(J=\Delta p\) and \(J = F_{avg}\Delta t\), then \(F_{avg}=\frac{\Delta p}{\Delta t}\). Given \(\Delta t = 1.54\ s\), \(F_{avg}=\frac{1.644}{1.54}\approx1.07\ N\).

Answer:

\(1.07\), \(N\)