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QJEEM23Ph92ea: JEE Main 2023 Physics - Class 11 JEEM23Ph92ea Solution: 100 balls each of mass moving with speed simultaneously strike…

100 balls each of mass m\mathrm{m} moving with speed vv simultaneously strike a wall normally and reflected back with same speed, in time t s\mathrm{t ~s} . The total force exerted by the balls on the wall is

Multiple Choice Options

  • A 200mvt\frac{200 m v}{t}
  • B 100mvt\frac{100 m v}{t}
  • C mv100t\frac{m v}{100 t}
  • D 200mvt200 m v t

Step-by-Step Verified Solution

When the balls strike the wall, the change in momentum of each ball is given by:

Δp=mv−(−mv)=2mv\Delta p = mv - (-mv) = 2mv

Since there are 100 balls, the total change in momentum of all the balls is ΔP=2m(100v)=200mv.\Delta P = 2m(100v) = 200mv. The time taken for all the balls to strike the wall is t\mathrm{t} seconds. Therefore, the average force exerted on the wall is given by:

F=ΔPt=2m(100v)t=200mvtF = \frac{\Delta P}{\mathrm{t}} = \frac{2m(100v)}{\mathrm{t}} = \frac{200mv}{\mathrm{t}}

Therefore, the total force exerted by the balls on the wall is F=200mvt\boxed{F = \frac{200mv}{\mathrm{t}}}

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