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MHT CET 2025 (5 May, Shift 2) · Impulse & Momentum Conservation

Impulse & Momentum Conservation question from MHT CET 2025 (5 May, Shift 2)

Q7MHT CET 2025 · 5 MAY · SHIFT 2Medium

Impulse & Momentum Conservation

A ball is dropped on the floor from a height of 20 m. It rebounds to a height of 5 m. Ball remains in contact with floor for 1 s. The average acceleration during contact is (acceleration due to gravity =10 m/s2= 10\text{ m/s}^{2})

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Answer: (a)

  1. Speed on hitting: v1=2gh1=2×10×20=20v_1 = \sqrt{2gh_1} = \sqrt{2 \times 10 \times 20} = 20 m/s (down).
  2. Speed on rebound: v2=2×10×5=10v_2 = \sqrt{2 \times 10 \times 5} = 10 m/s (up).
  3. Change in velocity =10−(−20)=30= 10 - (-20) = 30 m/s, so a=301=30 m/s2a = \frac{30}{1} = 30\text{ m/s}^2 (upward).
  4. So the answer is (a) 30 m/s230\text{ m/s}^2.
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