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 = 6160 N/m. A bullet of mass m = 8.40 g and velocity V overscript right arrow endscript of magnitude 700 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,

Mass of block, M = 5.4 kg

Spring constant, k = 6160 N/m

Mass of bullet, m = 8.40 g

Velocity, v = 700 m/s

(a) using law of conservation of momentum,

mi*vi = mf*vf

0.0084*700 = (5.40 + 0.0084)*vf

vf = 1.087 m/s

(b) KE = (1/2)*(M+m)*vf^2

KE = (1/2)* 5.4084*1.087^2

KE = 3.195 J

using law of conservation of energy,

PE = KE

(1/2)*k*x^2 = 3.195

(1/2)* 6160 *xmax^2 = 3.195

xmax = 0.0322 m

xmax = 3.22 cm

so amplitude, A = 3.22 cm

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