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A ball of mass 0.5 kg is dropped from a height of 10 m. The ball hits the ground and rises to a height of 1.5 m. The impulse imparted to the ball during its collision with the ground is : (Take \(g = 9.8 \, \text{m/s}^2\))
A ball of mass 0.5 kg is dropped from a height of 10 m. The ball hits the ground and rises to a height of 1.5 m. The impulse imparted to the ball during its collision with the ground is : (Take \(g = 9.8 \, \text{m/s}^2\))
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Impulse is the change in momentum: \(J = m(v_f - v_i)\). First, find the velocity just before impact (\(v_i\)) using \(v^2 = u^2 + 2gh\). Then, find the velocity just after impact (\(v_f\)) using the height of rebound and \(v^2 = u^2 + 2as\). Remember to consider the direction of velocities (upwards or downwards) when calculating the change in momentum.
Updated On: Nov 26, 2025
- \( 7 \text{ Ns} \)
- \( 0 \text{ Ns} \)
- \( 7 \sqrt{2} \text{ Ns} \)
- \( 21 \sqrt{2} \text{ Ns} \)
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The Correct Option is C
Solution and Explanation
Step 1: Impulse and Change in Momentum
Impulse equals the change in momentum: J = Δp = m (v_f - v_i), where m is mass, v_f is final velocity, and v_i is initial velocity.
Step 2: Velocity Just Before Impact
Using v² = u² + 2as, with u = 0, a = g = 9.8 m/s², and s = 10 m: v_i² = 0² + 2(9.8)(10) = 196, v_i = √196 = 14 m/s (downwards, so v_i = -14j).
Step 3: Velocity Just After Impact
The ball reaches a height of 1.5 m. At its peak, velocity is 0. Using v² = u² + 2as, with v = 0, a = -g = -9.8 m/s², and s = 1.5 m: 0² = v_f² + 2(-9.8)(1.5), v_f² = 29.4, v_f = √29.4 ≈ 5.42 m/s (upwards, so v_f = 5.42j).
Step 4: Impulse Calculation
Impulse calculation: J = m (v_f - v_i) = 0.5 (5.42j - (-14j)) = 0.5 (5.42 + 14)j, J = 0.5 × 19.42j = 9.71j Ns.
Conclusion:
The impulse imparted to the ball is approximately 9.71 Ns. The final answer is 9.71 Ns.
Impulse equals the change in momentum: J = Δp = m (v_f - v_i), where m is mass, v_f is final velocity, and v_i is initial velocity.
Step 2: Velocity Just Before Impact
Using v² = u² + 2as, with u = 0, a = g = 9.8 m/s², and s = 10 m: v_i² = 0² + 2(9.8)(10) = 196, v_i = √196 = 14 m/s (downwards, so v_i = -14j).
Step 3: Velocity Just After Impact
The ball reaches a height of 1.5 m. At its peak, velocity is 0. Using v² = u² + 2as, with v = 0, a = -g = -9.8 m/s², and s = 1.5 m: 0² = v_f² + 2(-9.8)(1.5), v_f² = 29.4, v_f = √29.4 ≈ 5.42 m/s (upwards, so v_f = 5.42j).
Step 4: Impulse Calculation
Impulse calculation: J = m (v_f - v_i) = 0.5 (5.42j - (-14j)) = 0.5 (5.42 + 14)j, J = 0.5 × 19.42j = 9.71j Ns.
Conclusion:
The impulse imparted to the ball is approximately 9.71 Ns. The final answer is 9.71 Ns.
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