A bullet of mass 10 g travelling horizontally with a velocity of 150 m s-strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Answers
Given parameters,
Initial velocity (u) = 150 m/s
Final velocity (v) = 0 (since the bullet finally comes to rest)
Time taken to come to rest (t) = 0.03 s
Find out
The distance of penetration of the bullet into the block. Also the magnitude of the force exerted by the wooden block on the bullet.
Solution
According to the first equation of motion,
v= u + at
Acceleration of the bullet (a)
0 = 150 + (a — 0.03 s)
a = -150 / 0.03
a = – 5000 m/s2
Here the negative sign indicates that the velocity of the bullet is decreasing.
According to the third equation of motion,
v2= u2+ 2as
0 = (150)2+ 2 (-5000)
s = 22500 / 10000
s = 2.25 m
Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion,
Force (F) = Mass × Acceleration
Mass of the bullet (m) = 10 g
m = 0.01 kg
Acceleration of the bullet (a) = 5000 m/s2
F = m × a
F = 0.01× -5000
F = -50 N
Hence, the magnitude of the force exerted by the wooden block on the bullet is 50 N.
The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object's speed and direction of motion (e.g. 60 km/h to the north).