A block of mass M = 5.40 kg, at rest on a horizontal frictionless table, is atta

Question

A block of mass M = 5.40 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 6120 N/m. A bullet of mass m = 8.20 g and velocity ModifyingAbove v With right-arrow of magnitude 500 m/s strikes and is embedded in the block (the figure). Assuming the compression of the spring is negligible until the bullet is embedded, determine (a) the speed of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.

Explanation / Answer

here,

mass of block, m = 5.40 kg

spring constant, k = 6120 N/m

mass of bullet, mb = 8.20g = 0.00820 kg
velocity of bullet, vb = 500 m/s

From Conservation of momentum, intital = final
mb*vb = (m+mb)*v

Solving for velocity of block after collision, v = mb*vb/(m+mb)
v = 0.00820*500/(0.00820+5.40)
v = 0.758 m/s

From Conservation of energy
KE loss by block = potential Energy gained by spring
0.5*(m+mb)*v^2 = 0.5*k*x^2

Solving for amplitude, x = sqrt( (m+mb)*v^2/k )
x = sqrt( (0.00820+5.40)*(0.758)^2/(6120) )
x = 0.023 m

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