Hello im kinda new here and hoping someone can help me with this energy conservation question (I believe).

Question

A 2.10g- bullet embeds itself in a 1.30kg - block, which is attached to a spring of force constant 770 N/m .

1) If the maximum compression of the spring is 5.60 cm , find the initial speed of the bullet.
2)Find the time for the bullet-block system to come to rest?

Thank you for any assistance

Answer 1

1) The embedding process is inelastic, so you have to use conservation of momentum to get the final velocity after the bullet is embedded and before spring compression starts. Let
V = initial bullet velocity,
v = velocity after it gets embedded
m = bullet mass
M = block mass
Momentum ocnservation tells you that
V*m = v*(m + M)
v = V*m/(m+M) = 1.613*10^-3 V
Spring compression is an elastic process, so the kinetic energy after embeddig = potential energy of thew fully compressed spring.
(1/2) kX^2 = (1/2)(m+M)v^2
= (1/2)m^2*V^2/(m+M)
Solve for V. Make sure X is in units of meters.

(2) The bullet-block system will keep on oscillating, so I think they want the time it takes for the spring to reach maximum compression (before it bounces back). That would be 1/4 of the period of oscillation, or
(1/4)*(2 pi)sqrt[k/(m+M)]

Answer 2

I agree with the previous post except for the last part, it would be (1/4)*(2 pi)sqrt[(m+M)/k] instead.