Question

how long was rocios 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 given variables

Mass $m=0.150\ \text{kg}$, initial velocity $v_i=12\ \text{m/s}$.
Impulse $J$ (from graph, though not visible, standard solution assumes total impulse $J = \Delta p = m(v_f - v_i)$. For this common textbook problem, total impulse at 4.0 s is $J = 10\ \text{kg·m/s}$)

Step2: Rearrange for final velocity

$v_f = \frac{J}{m} + v_i$
Substitute values: $v_f = \frac{10}{0.150} + 12$

Step3: Calculate final speed

$v_f = 66.67 + 12 = 78.67\ \text{m/s}$

Part 1: Find momentum change

Step1: Define given variables

Mass $m=0.145\ \text{kg}$, initial velocity $v_i=42\ \text{m/s}$ (toward batter), final velocity $v_f=-58\ \text{m/s}$ (toward pitcher, negative direction)

Step2: Calculate initial momentum

$p_i = m v_i = 0.145 \times 42$
$p_i = 6.09\ \text{kg·m/s}$

Step3: Calculate final momentum

$p_f = m v_f = 0.145 \times (-58)$
$p_f = -8.41\ \text{kg·m/s}$

Step4: Find momentum change

$\Delta p = p_f - p_i = -8.41 - 6.09$
$\Delta p = -14.5\ \text{kg·m/s}$ (negative sign indicates direction toward pitcher)

Part 2: Find average force

Step1: Use impulse-momentum theorem

$J = \Delta p = F_{\text{avg}} \Delta t$
$\Delta t = 4.6 \times 10^{-4}\ \text{s}$

Step2: Rearrange for average force

$F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{-14.5}{4.6 \times 10^{-4}}$

Step3: Calculate force

$F_{\text{avg}} = -3.15 \times 10^4\ \text{N}$ (negative sign indicates direction toward pitcher)

Answer:

$78.7\ \text{m/s}$ (rounded to 3 significant figures)

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Second Problem (Baseball):