Question

A 23.0kg block at rest on a horizontal frictionless table is connected to the wall via a spring with a spring constant k=31.0N/m. A 2.30×10-2kg bullet travelling with a speed of 540m/s embeds itself in the block. What is the amplitude of the resulting simple harmonic motion? Recall that the amplitude is the maximum displacement from equilibrium.

Answer #1

here,

mass of block , m1 = 23 kg

spring constant , K = 31 N/m

mass of bullet , m2 = 0.023 kg

initial speed , u = 540 m/s

let the final speed after the collison be v

using conservation of momentum for momentum

m2 * u = ( m1+ m2) * v

0.023 * 540 = ( 23 + 0.023) * v

solving for v

v = 0.54 m/s

let the amplitude of the resulting simple harmonic motion be x

using conservation of energy

0.5 * K * x^2 = 0.5 * (m1+ m2) * v^2

31 * x^2 = (23 + 0.023) * 0.54^2

x = 0.47 m

the amplitude of the resulting simple harmonic motion is 0.47 m

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