Question

1. this graph shows the velocity of a 30.0 kg dog running to catch a frisbee.

a. find the momentum of the dog while he was running at top speed.
b. find the total average impulse for the 16 seconds shown on the graph.
c. find the average impulse between 0 and 2 seconds.

Explanation:

Step1: Determine top - speed from the graph

From the velocity - time graph, the top - speed \(v = 4\space m/s\).

Step2: Calculate momentum

The formula for momentum is \(p=mv\). Given \(m = 30.0\space kg\) and \(v = 4\space m/s\), then \(p=(30.0\space kg)\times(4\space m/s)\)

Step1: Use the impulse - momentum theorem \(J=\Delta p=m(v_f - v_i)\)

From the graph, \(v_i = 0\space m/s\) (at \(t = 0\)) and at \(t = 16\space s\), assume the final velocity \(v_f\) (by counting the grid, if each small grid in velocity is \(0.5\space m/s\) and in time is \(0.5\space s\)).
The area under the velocity - time graph gives the displacement, but for impulse \(J=m(v_f - v_i)\).
If we assume the final velocity \(v_f=2.5\space m/s\) (by visual inspection of the end - point of the graph at \(t = 16\space s\)) and \(v_i = 0\space m/s\), \(m = 30\space kg\)
\(J=(30\space kg)\times(2.5\space m/s- 0\space m/s)\)

Step1: Use the impulse - momentum theorem \(J=\Delta p=m(v_f - v_i)\)

At \(t = 0\), \(v_i = 0\space m/s\). At \(t = 2\space s\), \(v_f = 4\space m/s\) (from the graph), \(m = 30\space kg\)
\(J=(30\space kg)\times(4\space m/s - 0\space m/s)\)

Answer:

\(120\space kg\cdot m/s\)