Question

A 4.8 kg block is suspended from a spring with a spring constant of 500 N/m. A 50 g bullet is fired into the block from directly below with a speed of 130 m/s and is embedded in the block. (a) Find the amplitude of the resulting simple harmonic motion. m (b) What fraction of the original kinetic energy of the bullet appears as mechanical energy in the harmonic oscillator?

Answer 1

Answer:

The fraction of the original kinetic energy of the bullet appears as mechanical energy in the harmonic oscillator is 0.00468.

Explanation:

Given that,

Mass of block = 4.8 kg

Mass of bullet = 50 g

Spring constant = 500 N/m

Speed of bullet = 130 m/s

(a). We need to calculate the velocity

Using conservation of momentum

$m_{b}v_{b}=(m+m_{b})v$

$v=\dfrac{m_{b}v_{b}}{(m+m_{b})}$

Put the value into the formula

$v=\dfrac{0.05\times130}{4.8+0.05}$

$v=1.340\ m/s$

We need to calculate the amplitude of the resulting simple harmonic motion

Using formula energy

$\dfrac{1}{2}(m+m_{b})v_{1}^2=\dfrac{1}{2}kx^2$

Put the value into the formula

$(4.8+0.05)\times1.340=500x^2$

$x^2=\dfrac{4.8+0.05}{500}$

$x=\sqrt{\dfrac{4.8+0.05}{500}}$

$x=0.0984\ m$

(b). We need to calculate the kinetic energy

Using formula of kinetic energy

$K.E=\dfrac{1}{2}\times0.05\times(130)^2$

$K.E=0.0423\ J$

$K.E=\dfrac{1}{2}mv^2+mgh$

for harmonic oscillator,

$K.E=\dfrac{1}{2}\times(4.8+0.05)\times(1.340)^2+(4.8+0.05)\times9.8\times0.0984$

$K.E=9.031\ J$

The fraction of the original kinetic energy of the bullet appears as mechanical energy in the harmonic oscillator

$\dfrac{K.E_{b}}{K.E}=\dfrac{0.0423}{9.031}$

$\dfrac{K.E_{b}}{K.E}=0.00468$

Hence, The fraction of the original kinetic energy of the bullet appears as mechanical energy in the harmonic oscillator is 0.00468.