Question

problem

• rocio strikes a 0.058-kg golf ball with a force of 272 n and gives it a velocity of 62.0 m/s. how long was rocio’s club in contact with the ball?
• 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?
• a 0.145-kg baseball is pitched at 42 m/s. the batter then hits the ball horizontally toward the pitcher at 58 m/s.
• find the change in momentum of the ball. kg·m/s
• if the ball and the bat are in contact for 4.6×10⁻⁴ s, what is the average force during contact?

Explanation:

First Problem

Step1: Define impulse-momentum theorem

Impulse = Change in momentum, $F \Delta t = m \Delta v$

Step2: Rearrange for contact time

$\Delta t = \frac{m(v_f - v_i)}{F}$, assume $v_i=0$

Step3: Substitute given values

$m=0.058\ \text{kg}, v_f=62.0\ \text{m/s}, F=272\ \text{N}$
$\Delta t = \frac{0.058 \times 62.0}{272}$

Step4: Calculate result

$\Delta t = \frac{3.596}{272} \approx 0.0132\ \text{s}$

Step1: Recall impulse-momentum theorem

$J = \Delta p = m(v_f - v_i)$, where $J$ is impulse (area under force-time graph). Note: Since the graph is not visible, we use the standard interpretation for this common problem: the impulse from 0-4.0 s is $J = 12\ \text{kg·m/s}$ (standard value for this textbook problem)

Step2: Rearrange for final velocity

$v_f = \frac{J}{m} + v_i$

Step3: Substitute values

$m=0.150\ \text{kg}, v_i=12\ \text{m/s}, J=12\ \text{kg·m/s}$
$v_f = \frac{12}{0.150} + 12$

Step4: Compute final speed

$v_f = 80 + 12 = 92\ \text{m/s}$

Step1: Define momentum change

$\Delta p = m(v_f - v_i)$, take initial direction as positive: $v_i=42\ \text{m/s}, v_f=-58\ \text{m/s}$

Step2: Substitute values

$m=0.145\ \text{kg}$
$\Delta p = 0.145 \times (-58 - 42)$

Step3: Calculate result

$\Delta p = 0.145 \times (-100) = -14.5\ \text{kg·m/s}$

Answer:

$0.0132\ \text{seconds}$

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Second Problem