Question

4. 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 second. 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.
3 Marks​

Answer 1

Answer:

50 N

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.

Hope it helps!

Have a great Day!

Answer 2

Explanation:

mass of bullet =10g

velocity = 150 m/s

time t =0.03s

force, f=m×a

a = change in velocity /time taken

a = 150/ 0.03

=5000 m/s2

gram covert to kg

= 10/1000=0.01kg

f= 0.01×5000= 50 kg m/s2 or 50N