wiz-icon
MyQuestionIcon
MyQuestionIcon
4
You visited us 4 times! Enjoying our articles? Unlock Full Access!
Question

Question 14
A bullet of mass 10 g travelling horizontally with a velocity of 150 ms1 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.

Open in App
Solution

Given, initial velocity, u = 150 ms1, final velocity, v=0 (since the bullet finally comes to rest), mass of the bullet, m=10 g=0.01 kg and time taken to come to rest, t=0.03 s.
Let the acceleration of a bullet be a.
We use first equation of motion, v=u+at,
0=150+a×0.03
a = 1500.03=5000 ms2
(Negative sign indicates that the velocity of the bullet is decreasing.)
Now we use third equation of motion:
v2=u2+2as,
0=1502+2×(5000)s
s=2250010000=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=m×a=0.01×5000=50 N
Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.

flag
Suggest Corrections
thumbs-up
11
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon