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Work Power and Energy question

2021 · 17 Mar · Shift 1 · Q43
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  5. /2021 · 17 Mar · Shift 1 · Q43

Work Power and Energy question

2021 · 17 Mar · Shift 1 · Q43

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A boy is rolling a 0.5 kg ball on the frictionless floor with the speed of 20 ms-1. The ball gets deflected by an obstacle on the way. After deflection it moves with 5% of its initial kinetic energy. What is the speed of the ball now?
  1. A
    14.41 ms −-− 1
  2. B
    19.0 ms −-− 1
  3. C
    4.47 ms −-− 1
  4. D
    1.00 ms −-− 1
View written solutionFree

Correct answer: C

  1. Given data

    • Mass of ball: m=0.5 kgm = 0.5\,\text{kg}m=0.5kg
    • Initial speed: u=20 m s−1u = 20\,\text{m s}^{-1}u=20m s−1
    • Final kinetic energy is 5%5\%5% of initial kinetic energy.
  2. Initial kinetic energy Ki=12mu2K_i = \frac{1}{2}mu^2Ki​=21​mu2

  3. Final kinetic energy Since the final kinetic energy is 5%5\%5% of the initial kinetic energy, Kf=0.05KiK_f = 0.05 K_iKf​=0.05Ki​

    Also, Kf=12mv2K_f = \frac{1}{2}mv^2Kf​=21​mv2

  4. Relate the speeds using kinetic energy 12mv2=0.05(12mu2)\frac{1}{2}mv^2 = 0.05\left(\frac{1}{2}mu^2\right)21​mv2=0.05(21​mu2)

    Cancelling 12m\frac{1}{2}m21​m from both sides, v2=0.05u2v^2 = 0.05u^2v2=0.05u2

    v=u0.05v = u\sqrt{0.05}v=u0.05​

  5. Substitute u=20 m s−1u = 20\,\text{m s}^{-1}u=20m s−1 v=200.05v = 20\sqrt{0.05}v=200.05​

    Since, 0.05=1200.05 = \frac{1}{20}0.05=201​ v=20120=2020=20×20×120v = 20\sqrt{\frac{1}{20}} = \frac{20}{\sqrt{20}} = \sqrt{20} \times \sqrt{20} \times \frac{1}{\sqrt{20}}v=20201​​=20​20​=20​×20​×20​1​ More directly, v=20×0.2236=4.472 m s−1v = 20 \times 0.2236 = 4.472\,\text{m s}^{-1}v=20×0.2236=4.472m s−1

    So, v≈4.47 m s−1v \approx 4.47\,\text{m s}^{-1}v≈4.47m s−1

  6. Check options

    • A: 14.41 m s−114.41\,\text{m s}^{-1}14.41m s−1 — incorrect
    • B: 19.0 m s−119.0\,\text{m s}^{-1}19.0m s−1 — incorrect
    • C: 4.47 m s−14.47\,\text{m s}^{-1}4.47m s−1 — correct
    • D: 1.00 m s−11.00\,\text{m s}^{-1}1.00m s−1 — incorrect

Final Answer: Option C is correct.

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