A 5.00-g bullet moving with an initial speed of v i = 390 m/s is fired into and

Question

A 5.00-g bullet moving with an initial speed of vi = 390 m/s is fired into and passes through a 1.00-kg block as shown in the figure below. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 930 N/m. The block moves d = 5.60 cm to the right after impact before being brought to rest by the spring.

(a) Find the speed at which the bullet emerges from the block.
m/s

(b) Find the amount of initial kinetic energy of the bullet that is converted into internal energy in bullet–block system during the collision.
J

Explanation / Answer

m =5 g, vi =390 m/s , M =1 kg , k =930 N/m , d=5.6 cm

from conservation of energy

(1/2)MV^2 = (1/2)kd^2

(1*V^2) = (930*5.6*5.6*10^-4)

V =1.708 m/s

from conservation of momentum

mvi +0= mvf +MV

(0.005*390) = (0.005*vf) +(1*1.708)

vf = 48.4 m/s

(b) Let the mechanical energy converted into internal energy in the collision is E Jules

=>E = KE(bullet) initial - KE(block) final - KE(bullet) final

= (0.5*0.005*390*390) - (0.5*0.005*48.4*48.4) -(0.5*1*1.708*1.708)

E =373 J

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