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Question
- a tennis ball with a mass of 0.025 kg is moving from left to right and 25 m/s. it is hit by a squash racket, which applies a force for 0.005 s so that the ball leaves the racket at 75 m/s moving from right to left. use the impulse - momentum theorem to calculate the average force on the ball. \\(\circ - 50\\ n\\) \\(\circ 25\\ n\\) \\(\circ - 75\\ n\\) \\(\circ - 5\\ n\\)
Step1: Define given values
$m=0.025\ \text{kg}, v_i=2.5\ \text{m/s}, v_f=-7.5\ \text{m/s}, \Delta t=0.005\ \text{s}$
(Note: Right is positive, left is negative)
Step2: Calculate momentum change
$\Delta p = m(v_f - v_i) = 0.025(-7.5 - 2.5)$
$\Delta p = 0.025(-10) = -0.25\ \text{kg·m/s}$
Step3: Apply impulse-momentum theorem
$F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{-0.25}{0.005}$
$F_{\text{avg}} = -50\ \text{N}$
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-50 N