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Center of Mass question

2022 · 25 Jul · Shift 2 · Q48
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Center of Mass question

2022 · 25 Jul · Shift 2 · Q48

JEE MainPhysicsCenter of MassMCQ+4 / −1
Two billiard balls of mass 0.05 kg each moving in opposite directions with 10 ms −-− 1 collide and rebound with the same speed. If the time duration of contact is t = 0.005 s, then what is the force exerted on the ball due to each other?
  1. A
    100 N
  2. B
    200 N
  3. C
    300 N
  4. D
    400 N
View written solutionFree

Correct answer: B

  1. Given data
  • Mass of each billiard ball: m=0.05 kgm = 0.05\,\text{kg}m=0.05kg
  • Initial speed of each ball: u=10 m s−1u = 10\,\text{m s}^{-1}u=10m s−1
  • They move in opposite directions and after collision rebound with the same speed.
  • Time of contact: Δt=0.005 s\Delta t = 0.005\,\text{s}Δt=0.005s
  1. Find change in momentum of one ball

Take one ball moving initially to the right.

  • Initial velocity: u=+10 m s−1u = +10\,\text{m s}^{-1}u=+10m s−1
  • Final velocity after rebound: v=−10 m s−1v = -10\,\text{m s}^{-1}v=−10m s−1

So, change in momentum is

Δp=m(v−u)\Delta p = m(v-u)Δp=m(v−u) Δp=0.05(−10−10)\Delta p = 0.05(-10-10)Δp=0.05(−10−10) Δp=0.05(−20)=−1 kg m s−1\Delta p = 0.05(-20) = -1\,\text{kg m s}^{-1}Δp=0.05(−20)=−1kg m s−1

Magnitude of change in momentum:

∣Δp∣=1 kg m s−1|\Delta p| = 1\,\text{kg m s}^{-1}∣Δp∣=1kg m s−1
  1. Use impulse-momentum theorem

Average force during collision:

F=∣Δp∣ΔtF = \frac{|\Delta p|}{\Delta t}F=Δt∣Δp∣​ F=10.005F = \frac{1}{0.005}F=0.0051​ F=200 NF = 200\,\text{N}F=200N
  1. Match with options
  • A: 100 N100\,\text{N}100N
  • B: 200 N200\,\text{N}200N
  • C: 300 N300\,\text{N}300N
  • D: 400 N400\,\text{N}400N

So the correct option is:

B: 200 N\boxed{\text{B: } 200\,\text{N}}B: 200N​
  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.

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