Question

A bullet of mass 10 gm travelling horizontally with a velocity of 150 m/s strikes a wooden target and comes to rest in 0.03 sec .The magnitude of force exerted by the wooden block is

Answer 1

Data: mass of the bullet (m) = 10 g = (10/1000) kg =0.01 kg Initial velocity (u) = 150 m/s Final velocity (v) = 0 m/s Time taken (t) = 0.03 sec i. Calculation of the distance of penetrationof the bullet into the block:  Calculation of the distance of penetrationof the bullet into the block = Displacement Displacement = average velocity × time Displacement = [(u + v) / 2] × timeDisplacement = [(150 + 0) / 2] × 0.03Displacement = 2.25 m ii. Calculation of the magnitude of the force exerted by the wooden block on the bullet: the magnitude of the force exerted by the wooden block on the bullet = Retarding force Retarding force = mass × reatardation Retarding force = mass ×[ (Intitial velocity - Final velocity) / t ] Retarding force = 0.01 ×[ (150 - 0) / 0.03 ] N Retarding force = [150/3] N Retarding force = 50 N

Answer 2

F=ma
F = 10×10^-3 × v-u/t
=> 10^-2 × 0-150/ 3×10-2
=> -150/3 N
=> -50 N
therefore the wooden blocks applies a force of 50N against the direction of the bullet

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