A ball of mass 0.1kg,travelling at 20m/s,strikes the palm of a player's hand and is stopped in 0.1 second. What is the magnitude of force exerted by the ball on the hand?

10 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20 N

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

30 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 20 N
Given : $Δ t = 0.1$ s $m = 0.1$ kg $u = 20$ m/s $v = 0$ m/s
Magnitude of change in momentum $Δ P = | m (v - u) | = | 0.1 (0 - 20) | = - 2$ kg.m/s
Force exerted by the ball $F = \frac{Δ P}{Δ t} = \frac{- 2}{0.1} = - 20$ N (-ve sign shows force exerted by ball and hand is in opposite direction)

Suggest Corrections

Similar questions

Q. (a) Explain why, a cricket player moves his hands backwards while catching a fast cricket ball.
(b) A 150 g ball, travelling at 30 m/s, strikes the palm of a player's hand and is stopped in 0.05 second. Find the force exerted by the ball on the hand.

Q. A player caught a cricket ball of mass

150 g

moving at a rate of

20 - m / s

. The catching process is completed in

0.1

second. What is the force exerted by the ball on the hand of the player?

Q. A player caught a cricket ball of mass

150 g

moving at a rate of

20 m / s

. If the catching process is completed in

0.1 s

, the average force of the blow exerted by the ball on the hand of the player is equal to :

A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to

A player caught a cricket ball of mass 150 gm moving at a rate of 20 m/s. If the catching process be completed in 0.1 s, then the force of the blow exerted by the ball on the hands of the player is

Join BYJU'S Learning Program

A ball of mass 0.1kg,travelling at 20m/s,strikes the palm of a player's hand and is stopped in 0.1 second. What is the magnitude of force exerted by the ball on the hand?

The correct option is B 20 NGiven : Δt=0.1 s m=0.1 kg u=20 m/s v=0 m/sMagnitude of change in momentum ΔP=|m(v−u)|=|0.1(0−20)|=−2 kg.m/sForce exerted by the ball F=ΔPΔt=−20.1=−20 N (-ve sign shows force exerted by ball and hand is in opposite direction)