A 5.00-g bullet moving with an initial speed of v i = 380 m/s is fired into and

Question

A 5.00-g bullet moving with an initial speed of vi = 380 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 = 4.00 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

Given,

Mass of bullet, mb = 5 g

Initial speed, vi = 380 m/s

Mass of block, M = 1 kg

applying law of conservation of momentum.

mb vb1 = mb vb2 + mblock vblock

0.005(380) = 0.005 vb2 + (1) vblock ................(1)

to find the velocity of the block, using law of conservation of energy

(1/2)mv2 = (1/2)kx2

v2 = 930 X (.04)2

v = 1.22 m/s

Now we plug this velocity for the blocks velocity in equation (1)

0.005(380) = 0.005vb2 + 1.22

vb2 = 136 m/s

b) KE=(1/2)mbvb12-[(1/2)mbvb22+(1/2)mblockvbock2]

KE= 0.5(0.005) X 3802 - [0.5(.005)1362+ (.5)1.222]

KE= 314 joules

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