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Que. A metal plate weighing 200g is balanced in mid-air by throwing 40 balls per second vertically upwards from below. After collision,balls rebound with the same speed. What is the speed with which the balls strike the plate ? [Given mass of each ball = 200g and g = 10ms^-2]

Solution with Explanation :-

Here we have,

Mass of plate(M_p) = 200g

Number of balls throwing per second(n) = 40

and mass of each ball(M_b) = 200g

First we convert the unit of g (acceleration due to gravity) from SI unit to CGS unit

g = 10ms^-2 = 1000cms^-2

To balance the plate in mid air, the force applied by 40 balls is equal to the weight (mg) of the plate.

Let initial velocity be u and final velocity be v 

and it is given that after collision balls rebound with the same speed, therefore  v = -u

Force applied by the 40 balls in 1 second is

F₁  = M_b × n × (v-u).   

 = 200×40×(-u-u)

= 8000(-2u)

= -16000u

Where  F₁ is the force applied by the 40 balls

M_b is the mass of each ball and

n is the number of balls thrown  per second

F₂ = M_p × -g

= 200 × -1000

= -200000 

Where M_p is the mass of plate 

F₁ = F₂

= -16000u = -200000

= u = 200000/16000

= u = 12.5 cms^-1

Thus the initial velocity of the ball is 12.5 cms^-1

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