A block of mass M = 4.80 kg, at rest on a horizontal frictionless table, is atta

Question

A block of mass M = 4.80 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 5210 N/m. A bullet of mass m = 9.20 g and velocity ModifyingAbove v With right-arrow of magnitude 670 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. A block of mass M = 4.80 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 5210 N/m. A bullet of mass m = 9.20 g and velocity ModifyingAbove v With right-arrow of magnitude 670 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

Given that,

mass of block, M = 4.8 kg

k = 5210 N/m

mass of bullet, m = 9.20 g = 0.0092 kg

vi = 670 m/s

(a)

Final speed of bullet = speed of block after collision = v

From law of momentum conservation,

m*vi = (M+m)*v

0.0092*670 = (4.8 + 0.0092) * v

v = 1.28 m/s

speed of the block immediately after the collision = 1.28 m/s

(b)

From law of energy conservation,

kinetic energy of block after collision = potential energy of spring

(1/2)*(M+m)*v^2 = (1/2)kx^2

(1/2)*(4.8092)*(1.28)^2 = (1/2)*5210*x^2

x = 0.0389 m

Hence, amplitude of the resulting simple harmonic motion = 0.0389 m

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