Question

A bullet of mass 10 g travelling horizontally with a velocity of
150 m s-1 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.​

Answer 1

Explanation:

Initial velocity of the bullet, u=150 ms

−1

Final velocity of the bullet, v=0

Mass of the bullet, m=10 g=0.01 kg

Time taken by the bullet to come to rest, t=0.03 s

Let distance of penetration be s

v=u+at

0=150+a(0.03)

⟹a=−5000 ms

−2

(- sign shows retardation)

v

2

=u

2

+2as

0=150

2

+2×(−5000)×s

s=2.25 m

Magnitude of the force exerted by the wooden block, ∣F∣=m∣a∣

∣F∣=0.01×5000=50 N

Answer 2

Explanation:

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

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 = -5000 m/s2

(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)

= 22500 / 10000

= 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 = 0.01 kg

Acceleration of the bullet, a = 5000 m/s2

F = ma = 0.01 * 5000 = 50 N

Hence, the magnitude of the force exerted by the wooden block on the bullet is 50 N.