Question

A cricket ball of mass 100 gm and travelling at 90 kmh– 1 is stopped by a player by moving his hands backward

by 40 cm. Find (i) time in which the ball comes to rest, and (ii) force exerted by player.​

Answer 1

Hi,

Here is your answer

Given:

Mass of ball(m)=100gm=0.1kg

Initial velocity of ball(u)=90km/hr=25m/s

Final velocity of ball(v)=0m/s

Distace (s)=40cm=0.4m

To Find:

i)Time taken

ii)Force exerted

Solution:

i)

By the third equation of motion, we know that

v²-u²=2as

Substituting the given datas, we get

0-25²=2×a×0.4

-625/0.8=a

a= -781.25m/s²(Acceleration is negative since it is retardation.)

By the first equation of motion,

v=u+at

0=25+-781.25×t

t=0.03s

ii)

By Newton's second law of motion,

F=ma

Substitute the datas given

F=0.1×-781.25

F= -78.125 N

Notes:

The three equations of motion are

• v=u+at
• s=ut+1/2×a×t²
• v²-u²=2as

Hope this helps you.

Answer 2

Explanation:

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