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NEET PHYSICSWORK, ENERGY AND POWERMedium

Question

A ball is thrown vertically downwards from a height of 20 m20 \text{ m} with an initial velocity v0v_0. It collides with the ground, loses 50%50\% of its energy in a collision and rebounds to the same height. The initial velocity v0v_0 is: (Take g=10 m/s2g = 10 \text{ m/s}^2)

A

14 m/s

B

20 m/s

C

28 m/s

D

10 m/s

Step-by-Step Solution

  1. Initial Mechanical Energy (EiE_i): At the point of projection (height h=20 mh = 20 \text{ m}), the ball possesses both kinetic energy (due to initial velocity v0v_0) and potential energy. Ei=Ki+Ui=12mv02+mghE_i = K_i + U_i = \frac{1}{2}mv_0^2 + mgh
  2. Energy Loss during Collision: The problem states the ball loses 50%50\% of its energy upon impact. Therefore, the energy remaining just after the collision (EfE_f) is 50%50\% of the initial energy. Ef=0.5Ei=0.5(12mv02+mgh)E_f = 0.5 E_i = 0.5 \left( \frac{1}{2}mv_0^2 + mgh \right)
  3. Rebound Energy: The ball rebounds to the same height h=20 mh = 20 \text{ m}. At the maximum height of the rebound, velocity is zero, so the energy is entirely potential. Erebound=mghE_{rebound} = mgh
  4. Conservation of Energy (Post-Collision): The remaining energy after collision must equal the energy required to reach the rebound height. 0.5(12mv02+mgh)=mgh0.5 \left( \frac{1}{2}mv_0^2 + mgh \right) = mgh
  5. Solve for v0v_0: 12mv02+mgh=2mgh\frac{1}{2}mv_0^2 + mgh = 2mgh 12mv02=mgh\frac{1}{2}mv_0^2 = mgh v02=2ghv_0^2 = 2gh Substitute g=10 m/s2g = 10 \text{ m/s}^2 and h=20 mh = 20 \text{ m}: v0=2×10×20=400=20 m/sv_0 = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \text{ m/s}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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