Question

19. A metal plate weighing 200 g 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 =
and g = 10 m/s]
2009​

Answer 1

Answer:

Here we have,

Mass of plate(Mp) = 200g

Number of balls throwing per second(n) = 40

and mass of each ball(Mb) = 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 = Mb × n × (v-u).

= 200×40×(-u-u)

= 8000(-2u)

= -16000u

Where F1 is the force applied by the 40 balls

Mb is the mass of each ball and

n is the number of balls thrown per second

Step-by-step explanation:

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