Question

A cricket ball of mass 0.15kg is moving with a velocity of 1.5m/s. find the impulse on the average force applied by the player if he is able to stop the ball in 0.18s

Answer 1

mass=0.15kg
initial velocity=1.5m/s
final velocity=0m/s(as the ball stopped at last)
time=0.18sec
force = m×a
F=m×(v-u/t)
F=0.15×(0-1.5/0.18)
F=0.15×(-1.5/0.18)
F=0.15×(-150/18)
F=0.15×(-8.3333)
F=-1.25Newton


HOPE YOU UNDERSTAND!!! PLEASE LIKE MY ANSWER

Answer 2

Answer:

A cricket ball of mass 0.15kg is moving with a velocity of 1.5m/s.

• The impulse will be $0.225\ kgm/s$

• The force applied by the player to stop the ball in $0.18$ s  will be $1.25\ N$

Explanation:

According to Newton's second law of motion,

Force (F) is the product of mass(m) and acceleration(a). It  can be expressed as,

$F = ma$                ...(1)

We know the impulse (p) is the product of mass (m) and velocity (v).

Or

$p = mv$            ...(2)

Then the acceleration is the rate of change of velocity with respect to time. can be expressed as

$a = \frac{v}{t}$                     ...(3)

Substitute equation(3) in (1) and then in (2)

$F = m\times \frac{v}{t} =\frac{p}{t}$     ...(4)

In the question, it is given that,

$m = 0.15 \ kg$

$v = 1.5\ m/s$

$t = 0.18\ s$

Substitute these values into equation (2)

$p = 0.15\times1.5= 0.225\ kgm/s$

then, equation(4) becomes,

$F = \frac{0.225}{0.18} = 1.25\ N$

So,

The impulse =  $0.225\ kgm/s$

The force applied by the player in $0.18\ s$  $=\ 1.25\ N$