Question

A bullet of mass 25 g moving with an initial velocity 100m/s , strikes a wooden block and comes to rest after penetrating a distance 2cm in it. Find the (i) retardation of the bullet and (ii) resistive force exerted by the wooden block. [4]

Answer 1

u=100m/s

v=0m/s

The motion of the bullet before it strikes the wooden planck is not of any importance to us

S=0.02m

v^2=u^2+2aS

Therefore,

0=10000+2*0.02*a

Therefore

a=-2.5*10^5m/s^2

Therefore retardation=2.5*10^5m/s^2

Resistive Force=m*a=6.25*10^3N

Answer 2

Explanation:

ANSWER

Mass of the bullet is m

b

=50gm=0.05Kg

Initial velocity of bullet u

b

=100m/s

Final velocity is v

b

=0m/s

(1) Initial momentum of bullet p

i

=m

b

u

b

=5Kgm/s

(2) Final momentum of bullet p

f

=m

b

v

b

=0.05×0=0

(3) Retardation occurred by the wooden block is

v

2

−u

2

=2as

−(100)

2

=2a(0.02)

a=−250000m/s

2

(4) Resistive force

F=ma

=0.05×−250000

=12500N

I hope it's help you