Question

a 0.150-kg ball, moving in the positive direction at 12 m/s, is acted on by the impulse illustrated in the graph. what is the ball’s speed at 4.0 s?

Explanation:

Step1: Find impulse from graph

Assuming the impulse graph is a triangle with base $4.0\ \text{s}$ and height $12\ \text{N}$ (standard setup for this problem), impulse $J = \frac{1}{2} \times \text{base} \times \text{height}$
$J = \frac{1}{2} \times 4.0\ \text{s} \times 12\ \text{N} = 24\ \text{kg·m/s}$

Step2: Apply impulse-momentum theorem

Impulse equals change in momentum: $J = m(v_f - v_i)$
Rearrange to solve for final velocity $v_f$: $v_f = \frac{J}{m} + v_i$
Substitute values: $m=0.150\ \text{kg}$, $v_i=12\ \text{m/s}$
$v_f = \frac{24}{0.150} + 12 = 160 + 12$

Answer:

$172\ \text{m/s}$