Question

A ball weighing 10 gm hits a hard surface
vertically with a speed of 5 m/s and rebounds
with the same speed. The ball remain in contact
with the surface for 0.01 sec. The average force
exerted by the surface on the ball is :
(1) 100N (2) 10N (3) 1N (4) 150 N​

Answer 1

Answer:

options 2 is correct

hope it will help

Answer 2

Answer:

The average force exerted by the surface on the ball is 10 newton.

The correct answer is option no (2) 10 N.

Explanation:

Given,

m = 10 gm = 10/1000 kg = 0.01 kg

v1 = 5m/s

v2 = -5 m/s (it is given that it rebounds

reboundswith the same speed , it rebounds means that direction of velocity is opposite).

t = 0.01 sec

F = (∆P)/∆t

F = m (∆v)/∆t

F = m (v1 - v2)/∆t

F = 0.01 [5 - (-5)]/0.01 newton.

F = (0.01 × 10)/0.01 newton

F = 10 newton

Here

m is the mass of ball.

v1 is the velocity of ball before rebounds.

v2 is the velocity of ball after rebounds.

t is the time .

F is the force of ball.

∆P is the change in momentum.

∆v is the change in velocity.

Hence,

The average force exerted by the surface on the ball is 10 newton.