Question

A bullet of mass 50 gram is moving with a velocity of 500mper second. It penetrates 10 cm into a still target and comes to rest. Calculate (a) the kinetic energy possessed by the bullet and (b) the retarding force offered by the target.

Answer 1

(A) Velocity of bullet = 500 m/s

Hence, K .E = 1/2 mv2= 1/2 x 0.05 x (500)2= 6250 J

(B)Mass of the bullet, m = 50 g = 0.05 kg

Initial velocity, u = 500 m/s

Distance travelled before stopping, s = 10 cm = 0.1 m

Using, v2 = u2 + 2as

= > 0 = 5002 + 2a(0.1)

= > a = -1.25 × 106 m/s2

Thus, the force is = ma = -6.25 × 104 N

The negative sign indicates the direction of force and acceleration are opposite to the direction of motion of thw bullet.

Answer 2

Answer:

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