Question

A 3.10 g bullet moving at 270 m/s enters and stops in an initially stationary 2.30 kg wooden block on a horizontal frictionless surface.

a) What's the speed of the bullet/ block combination?

b) What fraction of the bullet's kinetic energy was lost in this perfectly inelastic collision?

c)How much work was done in stopping the bullet?

Answer 1

Mass of the bullet m = 3.10 g = 3.10 * 10^-3 kg = 0.0031 kg

Velocity of the bullet v = 270 m/s

Mass of the wooden block M = 2.30 kg

Intial velocity of the block V¹ = 0

After collision block and bullet stick and move together

Whatever be the type of collision linear momentum is conserved,so

m* v + 0 = ( m + M ) * V where V is the common velocity of the block and bullet

m * v = ( m + M ) * V

0.0031 * 270 = ( 0.0031 + 2.30 ) * V .............. (1)

From (1) we can write V = 270 * 0.0031 / (0.0031 + 2.30)

V = 0.36 m/s

Fractional loss in kinetic energy of the system when second object is at rest i.e V¹= 0 is given by

K_F = M/( M + m)

= 2.30/(2.3031)

= 0.998

Work done is given by change in kinetic energy of the system W = K_fin - K_int

Intial kinetic energy of the system is given by K_int = 1/2 * m * v² + 0 = 1/2 * m * v²

= 1/2 * 0.0031 * (270)²

= 112.995 J

Final kinetic energy of the system is given by K_fin = 1/2 * (m + M) * V²

= 1/2 * (0.0031 + 2.30) *  (0.36)²

= 0.149 J

so, Work done in stopping the bullet is equal to W = 0.149 - 112.995 = -112.845 J