A bullet of mass 10g travelling horizontally with a velocity of 150m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the force.
Answers
Answer:
Initial velocity, u= 150 m/s
Final velocity, v= 0 (since the bullet finally comes to rest)
Time taken to come to rest, t= 0.03 s
According to the first equation of motion, v= u + at
Acceleration of the bullet, a
0 = 150 + (a — 0.03 s)a = -150 / 0.03 = -5000 m/s2
(Negative sign indicates that the velocity of the bullet is decreasing.)
According to the third equation of motion:
v2= u2+ 2as
0 = (150)2+ 2 (-5000)
= 22500 / 10000
= 2.25 m
Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Force, F = Mass * Acceleration
Mass of the bullet, m = 10 g = 0.01 kg
Acceleration of the bullet, a = 5000 m/s2
F = ma = 0.01 * 5000 = 50 N
Hence, the magnitude of the force exerted by the wooden block on the bullet is 50 N.
Answer:
Initial velocity, u= 150 m/s
Final velocity, v= 0 (since the bullet finally comes to rest)
Time taken to come to rest, t= 0.03 s
According to the first equation of motion, v= u + at
Acceleration of the bullet, a
0 = 150 + (a — 0.03 s)a = -150 / 0.03 = -5000 m/s2
(Negative sign indicates that the velocity of the bullet is decreasing.)
According to the third equation of motion:
v2= u2+ 2as
0 = (150)2+ 2 (-5000)
= 22500 / 10000
= 2.25 m
Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Force, F = Mass * Acceleration
Mass of the bullet, m = 10 g = 0.01 kg
Acceleration of the bullet, a = 5000 m/s2
F = ma = 0.01 * 5000 = 50 N
Hence, the magnitude of the force exerted by the wooden block on the bullet is 50 N.
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