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NEET PHYSICSMOTION IN A STRAIGHT LINEMedium

Question

A ball is dropped on the floor from a height of 10 m10 \text{ m}. It rebounds to a height of 2.5 m2.5 \text{ m}. If the ball is in contact with the floor for 0.01 sec0.01 \text{ sec}, the average acceleration during contact is:

A

2100 m/s² downwards

B

2100 m/s² upwards

C

1400 m/s²

D

700 m/s²

Step-by-Step Solution

  1. Velocity just before impact (viv_i): The ball falls freely from height h1=10 mh_1 = 10 \text{ m}. Using the kinematic equation v2=u2+2ghv^2 = u^2 + 2gh with initial velocity u=0u=0 and g=9.8 m/s2g=9.8 \text{ m/s}^2: vi=2gh1=2×9.8×10=196=14 m/sv_i = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14 \text{ m/s}. Taking the upward direction as positive, vi=14 m/sv_i = -14 \text{ m/s} (downwards) .
  2. Velocity just after rebound (vfv_f): The ball rises to a height h2=2.5 mh_2 = 2.5 \text{ m}. Using v2=u22ghv^2 = u^2 - 2gh where the final velocity at the peak is 0: 0=vf22(9.8)(2.5)    vf=49=7 m/s0 = v_f^2 - 2(9.8)(2.5) \implies v_f = \sqrt{49} = 7 \text{ m/s}. Since it is rebounding, vf=+7 m/sv_f = +7 \text{ m/s} (upwards) .
  3. Change in Velocity (Δv\Delta v): Δv=vfvi=(+7)(14)=21 m/s\Delta v = v_f - v_i = (+7) - (-14) = 21 \text{ m/s}.
  4. Average Acceleration: Average acceleration is the change in velocity divided by the time interval (Δt=0.01 s\Delta t = 0.01 \text{ s}). aavg=ΔvΔt=210.01=2100 m/s2a_{avg} = \frac{\Delta v}{\Delta t} = \frac{21}{0.01} = 2100 \text{ m/s}^2 .
  5. Direction: Since the change in velocity is positive, the acceleration is directed upwards.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A STRAIGHT LINE. 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.

PHYSICSMOTION IN A STRAIGHT LINEdroppedheightreboundsheightcontact

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