Question

A bullet of mss 10 g travelling horizontally with a velocity of 150m/s stirkes a stationary wooden block and comes to rest in 0.03 seconds. Calculate the distance of penetration of the bullet into the block. Calculate the magnitude of force exerted by the wooden block on the bullet

Answer 1

mass of the bullet (m) = 10 g = (10/1000) kg =0.01 kg

Initial velocity (u) = 150 m/s

Final velocity (v) = 0 m/s

Time taken (t) = 0.03 sec

i. Calculation of the distance of penetration of the bullet into the block:  

Calculation of the distance of penetration of the bullet into the block = Displacement

Displacement = average velocity × time

Displacement = [(u + v) / 2] × time

Displacement = [(150 + 0) / 2] × 0.03

Displacement = 2.25 m

ii. Calculation of the magnitude of the force exerted by the wooden block on the bullet:

the magnitude of the force exerted by the wooden block on the bullet = Retarding force

Retarding force = mass × retardation

Retarding force = mass ×[ (Initial velocity - Final velocity) / t ]

Retarding force = 0.01 ×[ (150 - 0) / 0.03 ] N

Retarding force = [150/3] N

Retarding force = 50 N

Answer 2

Let,
mass of bullet = m = 10g =0.010kg
initial velocity = u= 150 m/s
time interval = t =0.03 seconds
as,
bullet stops after time t, it's final velocity = v = 0m/s
We know,
v = u + at
0 = 150 + a × 0.03
a = -5000 m/s.s
Now,
v.v = u.u + 2as
0 = 150 × 150 +2×(-5000)×s
s = 2.25 m