A cricket ball of mass 100 g moving with a speed of 40 m/s is brought to rest by a player in 0.02s. Find the average force applied by the player.

15 N

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

200 N

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

5 N

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

6 N

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

Open in App

Solution

The correct option is B
200 N

Force applied = Mass*(rate of change in momentum with time)

Force = 0.1x $\frac{40}{0.02}$ = 200 N

Suggest Corrections

Similar questions

A cricket ball of mass $200 g$ moving with a speed of 40 $m s^{- 1}$ is brought to rest by a player in 0.04s. Calculate the following :
(i) change in momentum of the ball,
(ii) average force applied by the player. [2 MARKS]

Q. A cricket ball of mass 100 g is moving with speed of 25 m/s and is brought to rest in 0.025 s. find the average force applied.

Q. A cricket ball of mass

200 g m

moving with velocity

15 m / s

is brought to rest by a player in

0.05 s e c

. What is the impulse of the ball and average force exerted by player?

Q. A cricket ball of mass 100g moving with a velocity of 20m/s is brought to rest by a player in 0.05s the impulse of the ball is

Q. A cricket ball of mass

100 g

moving with a velocity of

30 m / s

is brought to rest by a player in

0.03 s e c

find Rate of change in momentum of ball

.

Join BYJU'S Learning Program

A cricket ball of mass 100 g moving with a speed of 40 m/s is brought to rest by a player in 0.02s. Find the average force applied by the player.

The correct option is B 200 N Force applied = Mass*(rate of change in momentum with time) Force = 0.1x400.02 = 200 N

The correct option is B
200 N

Force applied = Mass*(rate of change in momentum with time)

Force = 0.1x $\frac{40}{0.02}$ = 200 N