Question

A bullet of mass 0.1 kg moving horizontally with a velocity of 20 m/s strikes a stationary target and is brought to rest in 0.1 s find the impluse and average force of impact

Answer 1

IMPULSE =
$f \times \: change \: in \: ti me$
F=ma

a=20m/s

m=0.1 kg

$f = 0.1 \times 20$
$f = 2kg \: m \: per \: sec$

Impulse = 2 × 0.1 = 0.2

Answer 2

Given :

Mass of bullet , M = 0.1 kg .

Initial velocity , u = 20 m/s .

Final velocity , v = 0 m/s .

The bullet is brought to rest in time , t = 0.1 s .

To Find :

The impulse and average force of impact .

Solution :

We know magnitude of impulse is given by change in momentum .

$I=|P_f-P_i|\\\\I=|mv-mu|\\\\I=|m(v-u)|\\\\I=|0.1\times -20|\ kg\ m/s \\\\I=2\ kg\ m/s$

Now , average force is given by :

$F=\dfrac{I}{\Delta t}\\\\F=\dfrac{2}{0.1}\ N\\\\F=20\ N$

The impulse and average force of impact is 2 kg m/s and 20 N respectively.

Hence , this is the required solution .