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JIPMER 2019
Exam
JIPMER
Year
2019
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JIPMER 2019 · Physics · Center of Mass, Momentum and Collision

If a ball of mass 0.1 kg hits the ground from the height of 20 m...

If a ball of mass 0.1 kg hits the ground from the height of 20 m and bounce back to the same height, then find out the force exerted on the ball if the time of impact is 0.04 sec . (Take, g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^2 )

Options

  1. A100 N(+j^)100 \mathrm{~N}(+\hat{\mathbf{j}})
  2. B200 N(+j^)200 \mathrm{~N}(+\hat{\mathbf{j}})
  3. C100(−j^)N100(-\hat{\mathrm{j}}) \mathrm{N}
  4. D1000 N(j^1000 \mathrm{~N}(\hat{\mathrm{j}} )

Correct answer

A. 100 N(+j^)100 \mathrm{~N}(+\hat{\mathbf{j}})

Step-by-step solution

Here, mass of ball m=0.1 kgm=0.1 \mathrm{~kg} Velocity attained by the ball before hitting the  ground (v)=2gh(−j^)=2×10×20(−j^)=−20j^/s\begin{aligned} \text{ ground }(\mathrm{v}) & =\sqrt{2 \mathrm{gh}}(-\hat{\mathrm{j}}) \\ & =\sqrt{2 \times 10 \times 20}(-\hat{\mathrm{j}})=-20 \hat{\mathrm{j}} / \mathrm{s} \end{aligned} Velocity of ball when bounce back to the same height after hitting the ground, V′=−v=−(−20j^)=20j^ m/s∴V^{\prime}=-v=-(-20 \hat{j})=20 \hat{j} \mathrm{~m} / \mathrm{s}\therefore Change in velocity ΔV=V′−V=20j^−(−20j^)=40j^ m/s∴\Delta V=V^{\prime}-V=20 \hat{j}-(-20 \hat{j})=40 \hat{j} \mathrm{~m} / \mathrm{s} \therefore Force exerted on the ball f=ΔPΔt=mΔV0.04[ given Δt=0.04]=0.1×40j^0.04=100 N(j^)\begin{aligned} \mathrm{f} & =\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}=\frac{\mathrm{m} \Delta \mathrm{V}}{0.04}[\text{ given } \Delta \mathrm{t}=0.04] \\ & =\frac{0.1 \times 40 \hat{\mathrm{j}}}{0.04}=100 \mathrm{~N}(\hat{\mathrm{j}}) \end{aligned}
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