Question

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 IS 200g)

Answer 1

Given : Mass of Plate = 0.2 Kg
Mass of Each Ball = 0.2 Kg
No. of Ball thrown = 40
The plate is balanced so the force exterted by plate is equal to the force of 40 balls
= Mg = 40 x mv/t
= 0.2 x 10 = 40 x 0.2 x v/1(t=1 sec)
= v = 0.25 m/sec

Answer 2

Answer:

0.25m/s

Explanation:

GIVEN

a metal plate weight 200 g is balanced in the

mid air by throwing 40 balls per second.

TO FIND THE SPEED WITH WHICH THE BALLS STRIKE THE PLATE.

according to the question,

weight of plate = force exerted by 40 balls per

second.

weight of plate = w = mg

Force exerted by 40 balls per second is =

F = 40 X ma = 40 X mv / t [ where t = 1 seconds ]

F = 40 X ma = 40 X mv

Therefore,

acceleration = g = 10 m/s^2

mg = mv

Let,

mass of plate = 200g = 0.2 kg

mass of ball = 200g = 0.2 kg

0.2 X 10 = 40 X 0.2 X v

v = 10 / 40

v = 0.25 m/s

Therefore,

velocity of each ball = 0.25 m/s