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NVAEasyJEE 2023Velocity & Acceleration

JEE Physics 2023 Question with Solution

A tennis ball is dropped onto the floor from a height of 9.8 m9.8 \, \text{m}. It rebounds to a height of 5.0 m5.0 \, \text{m}. The ball comes in contact with the floor for 0.2 s0.2 \, \text{s}. The average acceleration during contact is _____ m/s2\text{m/s}^2. (Given g=10 m/s2g = 10 \, \text{m/s}^2)

Answer

Correct answer:120

Step-by-step solution

Standard Method

Given: The ball is dropped from height 9.8 m9.8 \, \text{m}, rebounds to height 5.0 m5.0 \, \text{m}, contact time is 0.2 s0.2 \, \text{s}, and g=10 m/s2g = 10 \, \text{m/s}^2.

Find: The average acceleration during contact.

First find the velocity just before impact:

v2=u2+2ghv^2 = u^2 + 2gh

With u=0u = 0 and h=9.8 mh = 9.8 \, \text{m},

v=2×10×9.8=196=14 m/sv = \sqrt{2 \times 10 \times 9.8} = \sqrt{196} = 14 \, \text{m/s}

Now find the velocity just after rebound:

v2=u2+2ghv^2 = u^2 + 2gh

With u=0u = 0 and h=5.0 mh = 5.0 \, \text{m},

v=2×10×5.0=100=10 m/sv = \sqrt{2 \times 10 \times 5.0} = \sqrt{100} = 10 \, \text{m/s}

The direction of velocity reverses during contact, so the total change in velocity is:

Δv=14+10=24 m/s\Delta v = 14 + 10 = 24 \, \text{m/s}

Average acceleration is:

a=ΔvΔt=240.2=120 m/s2a = \frac{\Delta v}{\Delta t} = \frac{24}{0.2} = 120 \, \text{m/s}^2

Therefore, the average acceleration is 120 m/s2120 \, \text{m/s}^2, so the numerical answer is 120.

Common mistakes

  • Taking the change in velocity as 14−1014 - 10 is incorrect because the ball reverses direction on rebounding. Use the magnitudes with direction change accounted for, so Δv=14+10\Delta v = 14 + 10.

  • Using the dropped height or rebound height directly in a=ΔvΔta = \frac{\Delta v}{\Delta t} is wrong because acceleration during contact depends on change in velocity over contact time, not displacement during flight.

  • Forgetting to compute the two speeds separately can lead to error. First find the speed before impact from 9.8 m9.8 \, \text{m}, then the speed after rebound from 5.0 m5.0 \, \text{m}.

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