Question

A bullet of mass 10gram travelling horizontally with a velocity 150 m/s ,strikes a stationary wooden block and comes to rest in 0.03 second. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force of the force exerted by the wooden block on the bullet. ( Use second eq. of motion to find displacement)

Answer 1

Answer:

50N

Explanation: initial velocity of the bullet , u= 150ms-1

final velocity of the bullet ;v =0    mass of the bullet ,m= 10g  = 0.01kg     time taken by bullet to come to rest , t= 0.03        let distance of penetration be s      v= u+at      0= 150+a[0.03]        a=  5000ms-2      v2 =u2 +2 as   0=[ 150]2 + 2×[_5000]× s     s=2.25 m          magnitude of the force exerted by the wooden block [f]=ma      [f] = 0.01 ×5000  =50N .