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.

Asked by Vishal kumar | 1 year ago |  242

##### Solution :-

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

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

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

Time period (t) = 0.03 s

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

As per the first motion equation , v = u + at

Therefore, a =- $$\frac{v-u}{t}= \frac{0-150}{0.03}\,ms^{-2}$$

Acceleration of the bullet after striking the wooden block is -5000 ms-2

Now, from the motion equation (v2 - u2 ) = 2as, the distance of penetration(s) can be calculated as follows:

S = $$\frac{v^2 - u^2}{2a}= \frac{0^2-(150^2)}{2(-5000)}meters=2.25\, meters$$

As per the second law of motion , F = ma

Therefore,force exterted by the wooden block on the bullet (F) = $$0.01kg\,\times\,(-5000\,ms^2)$$

= 50 N

This implies that the wooden block exerts a force of magnitude 50 N on the bullet in the direction that is opposite to the trajectory of the bullet.

Answered by Shivani Kumari | 1 year ago

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