a bullet of mass 10 gram travelling horizontally with a velocity of 150m/ s strikes of stationary wooden lock and comes to rest in 0.03sec.calculate the distance of penetration of a bullet into the block.also calculate the magnitude of the first exerted by wooden block on the bullet.
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Initial velocity of the bullet (u1) = 150m/s
final velocity of the bullet (v) =0
time taken (t)= 0.03
Acceleration (a)=(v-u)/t
= (150-0)/0.03
= 150/0.03
= 5000m/s^2
using seconf equation of motion
distance travelled (s) = ut + (1/2)at^2
=(0)(0.03) +(1/2)(5000)(0.03)(0.03)
= 2.25
so the distnce travelled by the bullet was 2.25m.
mass of the bullet = 10g
= 10/1000kg
Acceleration = 5000m/s^2
force = mass × acceleration
so the force applied by the bullet of the wooden block = (10/1000) × 5000 kg.m/s^2
= 5N
final velocity of the bullet (v) =0
time taken (t)= 0.03
Acceleration (a)=(v-u)/t
= (150-0)/0.03
= 150/0.03
= 5000m/s^2
using seconf equation of motion
distance travelled (s) = ut + (1/2)at^2
=(0)(0.03) +(1/2)(5000)(0.03)(0.03)
= 2.25
so the distnce travelled by the bullet was 2.25m.
mass of the bullet = 10g
= 10/1000kg
Acceleration = 5000m/s^2
force = mass × acceleration
so the force applied by the bullet of the wooden block = (10/1000) × 5000 kg.m/s^2
= 5N
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