Chapter - 2 Force and Laws of Motion

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Q
Force and Laws of Motion

Question:

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:

 

Given, mass of the bullet (m) = 10g  = 0.01 kg

Initial velocity of the bullet (u) = 150 m/s

Final velocity of the bullet (v) = 0 m/s

Time (t) = 0.03 s

To find the distance of penetration, the acceleration of the bullet must be calculated.

Using the first equation motion,

v = u + at

0 = 150 + a \times 0.03

-150 = a \times 0.03

a = \frac{-150}{0.03}

a = -5000 \ ms^{-2}

Acceleration of the bullet after striking the wooden block is  -5000 \ ms^{-2}.

Now, using the third equation of motion: v^2 - u^2 = 2as

the distance of penetration (s) can be calculated as follows:

(0)^2 - (150)^2 = 2\times -5000\times a

-22500 = -10000\times a

s= \frac{22500}{10000}

s = 2.25\ m

Now, for the force exerted by the wooden block on the bullet.

Using the second law of motion,F = ma

F = 0.01 kg \times - 5000\ ms^{-2}

F = -50\ N

Therefore the magnitude of the force exerted by the wooden block on the bullet is 50 N.


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