Question

1. an archer shoots two arrows at a target. both arrows have the same mass, and are shot with the same initial velocity. one of the arrows sticks into the target, and the other bounces off backwards. which arrow had a greater impulse? (hint: consider the change in velocity)

Explanation:

Step1: Recall the impulse - momentum theorem

Impulse \(J=\Delta p = m(v_f - v_i)\). Let the initial velocity of both arrows be \(v_i\) (towards the target) and mass be \(m\). For the arrow that sticks, \(v_f = 0\). So, \(\Delta p_1=m(0 - v_i)=-mv_i\). For the arrow that bounces, assume it bounces back with velocity \(v_f=-v_i\) (opposite direction). Then \(\Delta p_2=m(-v_i - v_i)=- 2mv_i\).

Step2: Compare the magnitudes of impulses

The magnitude of impulse for the stuck arrow is \(|J_1| = mv_i\), and for the bouncing arrow is \(|J_2|=2mv_i\). Since impulse is the change in momentum, and the force exerted on the arrow (and by Newton's third law, the force on the target) is related to the impulse (\(J = F\Delta t\), assuming \(\Delta t\) is the same for both collisions), the arrow that bounces causes a greater change in momentum.

Answer:

The one that bounces