Question

a. a cricket ball of mass 0.5 kg moving with a velocity of 10 m/s is brought to rest by player in 5 s. Find the force applied by the player to stop the ball.

b. A bullet of mass 0.4 kg is shot from a rifle with a speed 100 m/s. if the bullet penetrates through a 5 cm wall with a velocity of 20 m/s, what is the force exerted by the wall on the bullet?

Answer 1

a.
m=0.5 kg
u= 10m/s
t= 5sec
v=0 (because the ball come at rest after appling force by player)
a=?
F=?
v= u+ at
o= 10+ a×5
-10=5a
a=(-2)
f= ma
=0.5×2
=1N

Answer 2

Given :

Mass of ball , m = 0.5 kg .

Velocity of ball , u = 10 m/s .

Final velocity of ball , v = 0 m/s .

Ball is bought to rest in , t = 5 s .

To Find :

The force applied by the player to stop the ball.

Solution :

The deceleration of ball is given by :

$v=u+at\\\\a=\dfrac{v-u}{t}\\\\a=\dfrac{0-10}{5}\\\\a=-2\ m/s^2$

So , Force applied is given by :

$F = ma \\\\F = 0.5\times 2 \ N\\\\F=1 \ N$

Therefore , the force applied by the player to stop the ball is 1 N .

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