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

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

2 answers

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