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Question 14

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

Mass of bullet, m = 10 g = 10 / 1000 = 0.01 kg

Initial velocity, u = 150 m/s

Final velocity, v = 0

Time taken, t = 0.03 s

Using the first equation of motion

 v = u + at

0 = 150 + (a ×0.03 s)

(Negative sign indicates that the velocity of the bullet is decreasing.)

Now, using the third equation of motion:

v2 = u2 + 2as

0 = (150)2 + 2 (−5000) s

Hence, the distance of penetration of the bullet is 2.25 m.

We know that

Force, F = Mass × Acceleration

F = ma = 0.01 × 5000 = 50 N

Hence, the magnitude of force is 50 N.

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