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
c, a, b
b, a, c
a, c, b

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

At the start of region A, \(p = 0\). At the end of region A (at \(t = 3s\)), let the momentum be \(p_{A}\). So, \(\Delta p_{A}=p_{A}-0=p_{A}\)

Step3: Analyze region B

In region B, the momentum is constant. So, \(\Delta p_{B}=p - p=0\) (because \(p\) (initial momentum in region B) is equal to \(p\) (final momentum in region B))

Step4: Analyze region C

At the start of region C (at \(t = 5s\)), the momentum is \(p\) (same as the constant momentum in region B). At the end of region C (at \(t=7s\)), \(p = 0\). So, \(\Delta p_{C}=0 - p=-p\). The magnitude of \(\Delta p_{C}\) is \(|p|\)

Since \(|\Delta p_{A}| = |\Delta p_{C}|\) and \(|\Delta p_{B}|=0\)

Answer:

A and C (a tie), B