Q. 144.1( 135 Votes )

# 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 :

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

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

Time, t = 0.03 s(Given)

We know that,

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

0 = 150 + a × 0.03

0.03 a = - 150a = -

a = (- 5000 ms^{2})

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

We know that,

v^{2} = u^{2} + 2as

(0)^{2} = (150)^{2} + 2 × (-5000) × s

0 = 22500 – 10000 × s

10000 s = 22500s =

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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