Question:
A tennis ball of mass 150 g is moving at $20~ms^{-1}$. A racket strikes it, reversing its direction with a final speed of $30~ms^{-1}$. If the contact time is 0.02 s, then the magnitude of the force (in N) exerted by the racket is ________.
A tennis ball of mass 150 g is moving at $20~ms^{-1}$. A racket strikes it, reversing its direction with a final speed of $30~ms^{-1}$. If the contact time is 0.02 s, then the magnitude of the force (in N) exerted by the racket is ________.
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When a ball bounces back, change in velocity is the sum of speeds: $v_1 + v_2$.
Updated On: Jun 26, 2026
- 1.5 N
- 3.75 N
- 15 N
- 150 N
- 375 N
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The Correct Option is
Solution and Explanation
Step 1: Concept
Force is equal to the rate of change of momentum ($F = \frac{\Delta p}{\Delta t}$).
Step 2: Meaning
Change in momentum $\Delta p = m(v - u)$. Since direction is reversed, $u = 20~ms^{-1}$ and $v = -30~ms^{-1}$.
Step 3: Analysis
$\Delta p = 0.150 \text{ kg} \times (30 - (-20)) = 0.150 \times 50 = 7.5 \text{ kg}\cdot ms^{-1}$. $F = \frac{7.5}{0.02}$.
Step 4: Conclusion
$F = 375 \text{ N}$. Final Answer: (E)
Force is equal to the rate of change of momentum ($F = \frac{\Delta p}{\Delta t}$).
Step 2: Meaning
Change in momentum $\Delta p = m(v - u)$. Since direction is reversed, $u = 20~ms^{-1}$ and $v = -30~ms^{-1}$.
Step 3: Analysis
$\Delta p = 0.150 \text{ kg} \times (30 - (-20)) = 0.150 \times 50 = 7.5 \text{ kg}\cdot ms^{-1}$. $F = \frac{7.5}{0.02}$.
Step 4: Conclusion
$F = 375 \text{ N}$. Final Answer: (E)
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