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 910 N/m. The block moves d = 4.40 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

here,

mass of the bullet , mb = 0.005 kg

vi = 390 m/s

mass of block , M = 1 kg

spring constant , k = 910 N/m

d = 4.4 cm

d = 0.044 m

let the speed of the system after the collison be v

0.5 * (M + m ) * v^2 = 0.5 * k* x^2

(1.005) * v^2 = 0.5 * 910 * 0.044 ^ 2

v = 0.93 m/s

let the speed of bullet emerges from the block be v0

using conservation of momentum

m * vi = ( m + M)*v + m * v0

0.005 * 390 = 1.005 * 0.93 + 0.005 * v0

v0 = 203.07 m/s

speed of bullet when it emerges out is 203.07 m/s

(b)

the amount of initial kinetic energy of the bullet that is converted into internal energy in bullet–block system during the collision , E = 0.5 * mb * ( vi^2 - v0^2)

E = 0.5 * 0.005 * ( 390^2 - 203.07^2)

E = 277.16 J

the amount of initial kinetic energy of the bullet that is converted into internal energy in bullet–block system during the collision is 277.16 J

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