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Center of Mass question

2022 · 26 Jul · Shift 2 · Q42
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Center of Mass question

2022 · 26 Jul · Shift 2 · Q42

JEE MainPhysicsCenter of MassMCQ+4 / −1
A ball of mass 0.15 kg0.15 \mathrm{~kg}0.15 kg hits the wall with its initial speed of 12 ms−112 \mathrm{~ms}^{-1}12 ms−1 and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100 N100 \mathrm{~N}100 N, calculate the time duration of the contact of ball with the wall.
  1. A
    0.018 s
  2. B
    0.036 s
  3. C
    0.009 s
  4. D
    0.072 s
View written solutionFree

Correct answer: B

  1. Given data

    • Mass of ball: m=0.15 kgm = 0.15\,\text{kg}m=0.15kg
    • Initial speed toward wall: u=12 m s−1u = 12\,\text{m s}^{-1}u=12m s−1
    • Final speed after rebounding: v=12 m s−1v = 12\,\text{m s}^{-1}v=12m s−1 in the opposite direction
    • Force by wall: F=100 NF = 100\,\text{N}F=100N

  2. Choose direction convention Let the direction toward the wall be positive. Then, u=+12 m s−1,v=−12 m s−1u = +12\,\text{m s}^{-1}, \quad v = -12\,\text{m s}^{-1}u=+12m s−1,v=−12m s−1

  3. Change in momentum Δp=m(v−u)\Delta p = m(v-u)Δp=m(v−u) Δp=0.15(−12−12)\Delta p = 0.15(-12-12)Δp=0.15(−12−12) Δp=0.15(−24)=−3.6 kg m s−1\Delta p = 0.15(-24) = -3.6\,\text{kg m s}^{-1}Δp=0.15(−24)=−3.6kg m s−1

    Magnitude of change in momentum: ∣Δp∣=3.6 kg m s−1|\Delta p| = 3.6\,\text{kg m s}^{-1}∣Δp∣=3.6kg m s−1

  4. Use impulse-momentum theorem FΔt=∣Δp∣F\Delta t = |\Delta p|FΔt=∣Δp∣ 100⋅Δt=3.6100\cdot \Delta t = 3.6100⋅Δt=3.6 Δt=3.6100=0.036 s\Delta t = \frac{3.6}{100} = 0.036\,\text{s}Δt=1003.6​=0.036s

  5. Match with options 0.036 s\boxed{0.036\,\text{s}}0.036s​ So the correct option is B.

  6. Verification with stored answer Stored correct answer: B

    Our derived answer is also B, so they agree.

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