A bullet of mass A bullet of mass 10 gram travelling horizontally with a velocity of 18 50 M per second 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 extended by the wooden block on the bullet
Answers
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] × time
Displacement = [(150 + 0) / 2] × 0.03
Displacement = 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:-
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] × time
Displacement = [(150 + 0) / 2] × 0.03
Displacement = 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