Physics, asked by DemonicCrimin4, 1 month ago

A bullet of mass 10g traveling horizontally with a velocity of 150ms1 strikes a stationary wooden
block and comes to rest in 0.05 s. Calculate the distance of penetration of the bullet into the block. Also,
calculate the force exerted by the wooden block on the bullet

Answers

Answered by sach200709
1

Answer:

F= 50N

Explanation:

Given parameters,  

Initial velocity (u) = 150 m/s

Final velocity (v) = 0 (since the bullet finally comes to rest)

Time taken to come to rest (t) = 0.03 s  

Find out  

The distance of penetration of the bullet into the block. Also the magnitude of the force exerted by the wooden block on the bullet.  

Solution

According to the first equation of motion,  

v= u + at  

Acceleration of the bullet (a)  

0 = 150 + (a — 0.03 s)  

a = -150 / 0.03  

a = – 5000 m/s2  

Here the negative sign indicates that the velocity of the bullet is decreasing.  

According to the third equation of motion,  

v2= u2+ 2as  

0 = (150)2+ 2 (-5000)  

s = 22500 / 10000  

s = 2.25 m  

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

From Newton’s second law of motion,  

Force (F) = Mass × Acceleration  

Mass of the bullet (m) = 10 g  

m = 0.01 kg  

Acceleration of the bullet (a) = 5000 m/s2  

F = m × a  

F = 0.01× -5000  

F = -50 N  

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

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