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Center of Mass question

2023 · 31 Jan · Shift 1 · Q57
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  5. /2023 · 31 Jan · Shift 1 · Q57

Center of Mass question

2023 · 31 Jan · Shift 1 · Q57

JEE MainPhysicsCenter of MassMCQ+4 / −1
100 balls each of mass m\mathrm{m}m moving with speed vvv simultaneously strike a wall normally and reflected back with same speed, in time t s\mathrm{t ~s}t s. The total force exerted by the balls on the wall is
  1. A
    200mvt\frac{200 m v}{t}t200mv​
  2. B
    100mvt\frac{100 m v}{t}t100mv​
  3. C
    mv100t\frac{m v}{100 t}100tmv​
  4. D
    200mvt200 m v t200mvt
View written solutionFree

Correct answer: A

  1. Change in momentum of one ball

A ball of mass mmm strikes the wall normally with speed vvv and is reflected back with the same speed vvv.

Take the direction toward the wall as positive.

  • Initial momentum of one ball: pi=mvp_i = mvpi​=mv
  • Final momentum of one ball: pf=−mvp_f = -mvpf​=−mv

So, change in momentum of one ball is

Δp=pf−pi=(−mv)−(mv)=−2mv\Delta p = p_f - p_i = (-mv) - (mv) = -2mvΔp=pf​−pi​=(−mv)−(mv)=−2mv

Magnitude of change in momentum of one ball:

∣Δp∣=2mv|\Delta p| = 2mv∣Δp∣=2mv
  1. Change in momentum of 100 balls

Since there are 100100100 balls, total change in momentum magnitude is

100×2mv=200mv100 \times 2mv = 200mv100×2mv=200mv
  1. Force on the wall

Average force is rate of change of momentum:

F=ΔpΔtF = \frac{\Delta p}{\Delta t}F=ΔtΔp​

Here the total time is ttt, so

F=200mvtF = \frac{200mv}{t}F=t200mv​

This is the force exerted by the wall on the balls in magnitude, and by Newton's third law, the balls exert the same magnitude of force on the wall.

  1. Checking options
  • A: 200mvt\dfrac{200mv}{t}t200mv​ ✅
  • B: 100mvt\dfrac{100mv}{t}t100mv​ ❌
  • C: mv100t\dfrac{mv}{100t}100tmv​ ❌
  • D: 200mvt200mvt200mvt ❌

Therefore, the correct option is A.

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