Question

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Answer 1

a = - 5000
S = 2.25
F = - 50N

Answer 2

Answer:

50N

Explanation:

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.