Question

A 4.00 kgblock is suspended from a spring with k = 500 N/m.A 50.0 gbullet is fired into the block from directly below with a speed of 150 m/sand becomes embedded in the block. (a) Find the amplitude of the resulting SHM. (b) What percentage of the original kinetic energy of the bullet is transferred to mechanical energy of the oscillator?

Short answer

1. Amplitude of the resulting SHM is 0.166 m.
2. Percentage of the original kinetic energy of the bullet transferred to mechanical energy of the oscillator is 1.23%.

Step-by-step solution

The given data

• Mass of the block is, M = 4.00 kg.
• Mass of the bullet is, m = 50.0 or 0.50 kg.
• Spring constant is, k = 500 N.m.
• Velocity of bullet is, $v_0=150\mathrm{m}/\mathrm{s}$.

Understanding the concept of conservation laws in oscillations

By applying conservation of momentum, we can find the velocity of the bullet-block system just after the collision. By applying conservation of energy and using the velocity and spring force we can calculate the value of the amplitude of the resulting SHM. By using the kinetic energy formula and taking the ratio of kinetic energies after and before the collision, we can find the percentage of the original kinetic energy of the bullet transferred to the mechanical energy of the oscillator.

Formula:

The magnitude of force of the spring in oscillations, F = kx (i)

The conservation law of momemtum in elastic collision,${\mathrm{mv}}_0=(\mathrm{m}+\mathrm{M})\mathrm{v}$ (ii)

The law of conservation of energy, ${\left(KE\right)}_i+{\left(PE\right)}_i+{\left(KE\right)}_f+{\left(PE\right)}_f$ (iii)

(a) Calculation of the amplitude of the resulting SHM

Let’s consider the displacement of the system as x.

The initial position of the body using equation (i) due to spring elongation is given as:

$$\begin{gathered} x_i=\frac{Mg}{k}(\because F=Mg) \\ =\frac{4.00\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2}{500\mathrm{N}/\mathrm{m}} \\ =\frac{39.2}{500} \\ =0.078\mathrm{m} \end{gathered}$$

Similarly, after collision between block and bullet results SHM about its equilibrium position and this elongation is given using equation (i) as:

$$\begin{gathered} x_i=\frac{(m+M)g}{k}(\because F=(\mathrm{m}+\mathrm{M}\left)\mathrm{g}\right) \\ =\frac{(0.50\mathrm{kg}+4.00\mathrm{kg})\times 9.8\mathrm{m}/{\mathrm{s}}^2}{500\mathrm{N}/\mathrm{m}} \\ =\frac{(4.05\mathrm{kg})\times 9.8\mathrm{m}/{\mathrm{s}}^2}{500\mathrm{N}/\mathrm{m}} \\ =0.79\mathrm{m} \end{gathered}$$

According to conservation of momentum, the final velocity using equation (ii) after the inelastic collision is given as:

$$\begin{gathered} v=\frac{m{v}_0}{m+M} \\ =\frac{0.050\mathrm{kg}\times 150\mathrm{m}/\mathrm{s}}{0.50\mathrm{kg}+4.00\mathrm{kg}} \\ =\frac{7.5}{4.05} \\ =1.85\mathrm{m}/\mathrm{s} \end{gathered}$$

Just after the bullet is embedded at the initial position, the kinetic energy of the system is given as:

${\left(KE\right)}_i=\frac{1}{2}(m+M)v^2$

And the elastic potential energy of the system at the initial is given as:

${\left(PE\right)}_i=\frac{1}{2}k{x_i}^2$

When the block and bullet reach the highest point during their motion, the kinetic energy before collision is 0.

Ifis the amplitude of the motion, thenthe spring is compressed byso the elastic potential energy is given as:

$PE=\frac{1}{2}k{(A-x)}^2$

And the gravitational potential energy before collision for gravity is given as:

${\left(PE\right)}_g=(m+M)gA$

Now, according to conservation of mechanical energy using,

$$\begin{gathered} \frac{1}{2}(m+M)v^2+\frac{1}{2}k{x_i}^2=0+(m+M)gA+\frac{1}{2}{(A-x)}^2 \\ \frac{1}{2}(m+M)v^2+\frac{1}{2}k{x_i}^2=(m+M)gA+\frac{1}{2}k(A^2+x^2-2Ax) \\ (m+M)v^2+\frac{1}{2}k{x_i}^2=2(m+M)gA+k(A^2+x^2-2Ax) \\ (m+M)v^2+\frac{1}{2}k{x_i}^2=2(m+M)gA+k{A}^2+k{x}^2-2kAx \\ {A}^2=\frac{(m+M)v^2}{k}+x_i^2-x^2+\frac{Mg}{k}-\frac{2(m+M)Ag}{k}-2Ax \\ (x_i=\frac{Mg}{k}\mathrm{and}x=\frac{(m+M)g}{k}) \\ {A}^2+\left(\frac{2mg}{k}+\frac{2Mg}{k}+2\frac{(m+M)g}{k}\right)A=\frac{(m+M)v^2}{k}-{\left(\frac{(m+M)g}{k}\right)}^2+\frac{Mg}{k} \end{gathered}$$

Therefore,

$$\begin{gathered} A^2=\frac{(m+M)v^2}{k}-\frac{m{g}^2}{k^2}(m+2M) \\ A=\sqrt{\left[\frac{(m+M)v^2}{k}-\frac{m{g}^2}{k^2}(m+2M\right]} \\ =\sqrt{\left[\frac{(0.050\mathrm{kg}+4.00\mathrm{kg})\times {(1.85\mathrm{m}/\mathrm{s})}^2}{500}-\frac{0.50\mathrm{kg}\times {(9.8\mathrm{m}/{\mathrm{s}}^2)}^2}{{500}^2}(0.050\mathrm{kg}+(2\times 4.00\mathrm{kg}\left)\right)\right]} \\ =\sqrt{\left[\frac{(4.050)\times 1.{85}^2}{500}-\frac{0.050\times 9.8^2}{500^2}(8.050)\right]} \\ =\sqrt{\left[0.0277-0.0001546\right]} \\ =0.166\mathrm{m} \end{gathered}$$

Hence, the value of the amplitude is 0.166 m

(b) Calculation of original kinetic energy to mechanical energy

Before collision, the energy of the bullet is given by,

$$\begin{gathered} E_0=\frac{1}{2}m{v_0}^2 \\ =\frac{1}{2}\times 0.050\mathrm{kg}\times {(150\mathrm{m}/\mathrm{s})}^2 \\ =562.5\mathrm{J} \end{gathered}$$

Kinetic energy just after collision between block and bullet is given as:

$$\begin{gathered} E=\frac{1}{2}(m+M)v^2 \\ =\frac{1}{2}(0.050\mathrm{kg}+4.00\mathrm{kg})\times {(1.85\mathrm{m}/\mathrm{s})}^2 \\ =6.94\mathrm{J} \end{gathered}$$

Since, the block is at rest during the collision

Percentage of the original kinetic energy of the bullet is transferred to mechanical energy of the oscillator is given as:

$$\begin{gathered} \%Energytransferred=\frac{E}{E_0}\times 100 \\ =\frac{6.94J}{562.5J}\times 100 \\ =1.23\% \end{gathered}$$

Hence, the required value of the percentage is 1.23%