Question

An 8.6 g bullet is fired into and embeds itself in a 1.48 kg block attached to a spring with a spring constant of 20.4 N/m and whose mass is negligible. How far is the spring compressed if the bullet has a speed of 123 m/s just before it strikes the block, and the block slides on a frictionless surface? Answer in units of m.

Answer 1

To solve this problem it is necessary to apply the concepts concerning conservation of the linear momentum as well as the equations of energy acquisition for springs.

From the definition of conservation of linear momentum we have to,

`m_1u_1+m_2u_2 = (m_1+m_2)V_f`

Where,

`V_f =`Final velocity

`m_1 =`Mass of the bullet

`m_2 =`Mass of the block

`u_1 =` Initial velocity of the bullet

`u_2 =` initial velocity of the block

The block does not have initial speed because it is at rest, then replacing we have to,

`(8.6*10^(-3))(123)+(1.48)(0)=(8.6*10^(-3)+1.48)V_f`

Re-arrange to find
`V_f,`

`V_f = ((8.6*10^(-3))(123))/((8.6*10^(-3)+1.48))`

`V_f = 0.7106m/s`

Now applying the energy conservation equations we have that the potential and kinetic energy of the spring must be maintained in the way

`KE_i+PE_i = KE_f+PE_f`

If the spring is compressed, then the velocity becomes zero. Here the kinetic energy is zero and the spring potential energy as follow,

`0+(1)/(2)kx^2=(1)/(2)Mu^2+0`

Re-arrange to find x,

`x = v\sqrt{(M)/(k)}`

`x = 0.7106\sqrt{(1.48)/(20.4)}`

`x = 0.1913m`

Therefore the spring is compressed around to 0.1913m