Question:
A ball of $200g$ mass moving with a speed of $5\,ms^{-1}$ collides with a wall and bounces back with the same speed. If the force exerted on the wall is $1N$, then the ball is in contact with the wall for:
A ball of $200g$ mass moving with a speed of $5\,ms^{-1}$ collides with a wall and bounces back with the same speed. If the force exerted on the wall is $1N$, then the ball is in contact with the wall for:
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Reversal of velocity doubles momentum change.
Updated On: Apr 24, 2026
- $2s$
- $1s$
- $0.5s$
- $1.5s$
- $0.75s$
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The Correct Option is A
Solution and Explanation
Concept:
• Impulse = change in momentum = $F \cdot t$
Step 1: Convert mass
\[ m = 200g = 0.2\,kg \]
Step 2: Change in momentum
\[ \Delta p = m(v_f - v_i) = 0.2(-5 - 5) = -2 \] Magnitude: \[ |\Delta p| = 2 \]
Step 3: Use impulse
\[ F t = \Delta p \] \[ 1 \cdot t = 2 \Rightarrow t = 2s \] Final Conclusion:
\[ = 2s \]
• Impulse = change in momentum = $F \cdot t$
Step 1: Convert mass
\[ m = 200g = 0.2\,kg \]
Step 2: Change in momentum
\[ \Delta p = m(v_f - v_i) = 0.2(-5 - 5) = -2 \] Magnitude: \[ |\Delta p| = 2 \]
Step 3: Use impulse
\[ F t = \Delta p \] \[ 1 \cdot t = 2 \Rightarrow t = 2s \] Final Conclusion:
\[ = 2s \]
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