Question

a cricket ball of mass 150g is moving with a velocity of 12m/s and hitted by a bat so that the ball is turned back with a velocity of 20m/s. if the duration of contact btw the ball and bat is 0.01s.then:
1. the impulse is----------------------

Answer 1

The impulse of the ball is 4.8 Ns.

Given data:

Initially, velocity of the ball = 12 m/s  

Final velocity = 20 m/s

Mass, m = 150 g = 0.15 kg

To find:

Impulse = ?

Solution:

$\text {Impulse} =F \Delta t$

We know that,

$F=\frac{\Delta p}{\Delta t}$

Where,

p is momentum

Therefore,

$\therefore \text {Impulse}=\frac{\Delta p}{\Delta t} \times \Delta t$

$\Rightarrow \text {Impluse}=\Delta p$

$\Delta p=m v_{2}-m v_{1}=m\left(v_{1}-v_{2}\right)$

$\Delta p=0.15 \times(-20-12)$

Since, the "rebound velocity" is taken negative in motion as the direction is changed.

Thus,

Impulse = 4.8 Ns

Answer 2

Answer:

force = mass x distance

  or

f=ma

f=m x change in momentum/time

f=0.15 x [20m/s{-12m/s}] /0.01s

f=15 x32 = 480N