Question

A bullet of mass 10 g travelling horizontally with a velocity of 150 ms-1 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 magnitude of the force exerted by the wooden block on the bullet.​

Answer 1

Answer:

Initial velocity, u= 150 m/s

Initial velocity, u= 150 m/sFinal velocity, v= 0 (s comes to rest)

comes to rest)Time taken to come to rest, t= 0.03 s

comes to rest)Time taken to come to rest, t= 0.03 sAccording to the first equation of motion, v= u + at

comes to rest)Time taken to come to rest, t= 0.03 sAccording to the first equation of motion, v= u + atAcceleration of the bullet, a

comes to rest)Time taken to come to rest, t= 0.03 sAccording to the first equation of motion, v= u + atAcceleration of the bullet, a0 = 150 + (a — 0.03 s)a = -150 / 0.03 = -5000 m/s2

comes to rest)Time taken to come to rest, t= 0.03 sAccording to the first equation of motion, v= u + atAcceleration of the bullet, a0 = 150 + (a — 0.03 s)a = -150 / 0.03 = -5000 m/s2(Negative sign indicates that the velocity of the bullet is decreasing.

comes to rest)Time taken to come to rest, t= 0.03 sAccording to the first equation of motion, v= u + atAcceleration of the bullet, a0 = 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:

comes to rest)Time taken to come to rest, t= 0.03 sAccording to the first equation of motion, v= u + atAcceleration of the bullet, a0 = 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

comes to rest)Time taken to come to rest, t= 0.03 sAccording to the first equation of motion, v= u + atAcceleration of the bullet, a0 = 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+ 2as0 = (150)2+ 2 (-5000)

comes to rest)Time taken to come to rest, t= 0.03 sAccording to the first equation of motion, v= u + atAcceleration of the bullet, a0 = 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+ 2as0 = (150)2+ 2 (-5000)= 22500 / 10000

comes to rest)Time taken to come to rest, t= 0.03 sAccording to the first equation of motion, v= u + atAcceleration of the bullet, a0 = 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+ 2as0 = (150)2+ 2 (-5000)= 22500 / 10000= 2.25 m

comes to rest)Time taken to come to rest, t= 0.03 sAccording to the first equation of motion, v= u + atAcceleration of the bullet, a0 = 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+ 2as0 = (150)2+ 2 (-5000)= 22500 / 10000= 2.25 mHence, the distance of penetration of the bullet into the block is 2.25 m.

Answer will be 50Newton

Explanation:

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