Q26 of 49 Page 9

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.

Firstly, we have to calculate the acceleration of the bullet:

Initial velocity, u = 150 m/s (Given)

Final velocity, v = 0 (As the bullet comes to rest after striking the wooden block)

Time, t = 0.03 s(Given)

We know that,

v = u + at
Putting the Values in the equation,We get:-

0 = 150 + a × 0.03

0.03 a = - 150

a = -

a = (- 5000 ms2)

Now, we will calculate the distance of penetration of bullet:


We know that,


v2 = u2 + 2as


(0)2 = (150)2 + 2 × (-5000) × s


0 = 22500 – 10000 × s

10000 s = 22500

s =


s = 2.25 m


Now, we have to calculate the magnitude of the force:

We know that:-
Force, F = m × a


F = kg × (-5000) m/s (1 gram=1/1000 kg)


= -50 N

Thus the magnitude of retarding force is exerted by the wooden block on the bullet is 50 N

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