Question

a particle is moving along the +x axis, and the graph shows its momentum p as a function of time t. use the impulse - momentum theorem and rank (largest to smallest) the three regions according to the magnitude of the impulse applied to the particle. a and c (a tie), b a, b, c a, c, b c, a, b b, a, c

Explanation:

Step1: Recall the impulse - momentum theorem

The impulse - momentum theorem states that \(J=\Delta p\), where \(J\) is the impulse and \(\Delta p\) is the change in momentum.

Step2: Analyze region A

In region A, the momentum changes from \(p = 0\) to some non - zero value \(p_{A}\). Let the final momentum in region A be \(p_{A}\), so \(\Delta p_{A}=p_{A}-0 = p_{A}\)

Step3: Analyze region B

In region B, the momentum is constant. So, \(\Delta p_{B}=p - p=0\) (where \(p\) is the constant momentum value in region B)

Step4: Analyze region C

In region C, the momentum changes from a non - zero value (same as the final value in region A, say \(p_{C}\)) to \(p = 0\). So, \(\Delta p_{C}=0 - p_{C}\), and \(|\Delta p_{C}|=|p_{C}|\). Since the magnitude of the change in momentum in region A (\(|\Delta p_{A}|\)) and region C (\(|\Delta p_{C}|\)) is the same (because the graph has symmetric rise and fall in momentum for A and C) and \(|\Delta p_{B}| = 0\)

Answer:

A and C (a tie), B