A 20g bullet moving at 200 meter per second ts a bag of sand and comes to rest in 0.011(s) .what is the momentum of the bullet just before tting the bag ? find the average force that stopped the bullet.
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Given mass of bullet = 20g = 0.02kg
Given initial speed of bullet = 200m/s
and momentum = mass× velocity
so, momentum of the bullet would be = 0.02 × 200 kg.m/s
=4kg.m/s
final velocity = 0
time taken to come to rest = 0.011 s
acceleration = (final velocity- initial velocity)/ time
= (0-200)/0.011
= 200000/11 m/s^2
force = mass× acceleration
hence the force = (0.02)(200000/11)
= 4000/11
=363.63 N.
Given initial speed of bullet = 200m/s
and momentum = mass× velocity
so, momentum of the bullet would be = 0.02 × 200 kg.m/s
=4kg.m/s
final velocity = 0
time taken to come to rest = 0.011 s
acceleration = (final velocity- initial velocity)/ time
= (0-200)/0.011
= 200000/11 m/s^2
force = mass× acceleration
hence the force = (0.02)(200000/11)
= 4000/11
=363.63 N.
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