A 5.00-g bullet moving with an initial speed of v i = 440 m/s is fired into and

Question

A 5.00-g bullet moving with an initial speed of v i = 440 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 920 N/m. The block moves d = 4.00 cm to the right after impact before being brought to rest by the spring. Find the speed at which the bullet emerges from the block. m/s Find the amount of initial kinetic energy of the bullet that is converted into internal energy in bullet-block system during the collision.

Explanation / Answer

initial speed of bullet , vi = 440 m/s

mass of bullet , m = 0.005 Kg

mass of block , M = 1 Kg

let the speed of the block after collision is v

Using conservation of enegry

0.5 * k * d^2 = 0.5 * m * v^2

920 * 0.04^2 = 1 * v^2

v = 1.213 m/s

Using conservation of momentum

0.005 * (440 - vf) = 1.213 * 1

vf = 197 m/s

the final speed of bullet is 197 m/s

b)

amount of kinetic energy lost = 0.5 * 0.005 * 440^2 - 0.5 * 0.005 * 197^2 - 0.5 * 1 * 1.213^2

amount of kinetic energy lost = 386.24 J

the amount of kinetic energy lost is 386.24 J

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