Question

A 15.0 kg block is attached to a very light horizontal spring of force constant 400 N/m and is resting on a smooth horizontal table. (See the figure below .) Suddenly it is struck by a 3.00 kg stone traveling horizontally at 8.00 m/s to the right, whereupon the stone rebounds at 2.00 m/s horizontally to the left.

Find the maximum distance that the block will compress the spring in m after the collision.

Answer 1

Concepts and reason

The concept required to solve this question is the conservation of momentum and law of conservation of energy.

Initially, apply conservation of linear momentum. Then rearrange the expression for the velocity of the block after the collision.

After that, apply law of conservation of energy to find the distance that the block compresses.

Fundamentals

The conservation of the momentum states that during the collision of the particles, the initial momentum is equal to the final momentum. This can be represented as,

${p_{\rm{i}}} = {p_{\rm{f}}}$

Here, ${p_{\rm{i}}}$ is the initial momentum and ${p_{\rm{f}}}$ is the final momentum.

The expression of the momentum is,

$p = mv$

Here, p is the momentum, m is the mass, and v is the velocity.

The conservation of the energy states that during the collision of the particles, the initial energy is equal to the final energy. This can be represented as,

${E_{\rm{i}}} = {E_{\rm{f}}}$

Here, ${E_{\rm{i}}}$ and ${E_{\rm{f}}}$ are the initial and final energy of the system.

The expression of the kinetic energy is,

$KE = \frac{1}{2}m{v^2}$

Here, KE is the kinetic energy, m is the mass, and v is the velocity.

The potential energy U of the system with spring constant k is,

$U = \frac{1}{2}k{x^2}$

Here, x is the compression in the spring.

Let a ball hit a block attached by a spring to the wall.

The expression for the conservation of momentum is,

${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$

Here, ${m_1}$ is the mass of the block , ${u_1}$ is the initial velocity of the block , ${m_2}$ is the mass of the stone , ${u_2}$ is the initial velocity of the stone , ${v_1}$ is the final velocity of the block and ${v_2}$ is the final velocity of the stone .

Substitute 0 for ${u_1}$ in equation ${m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$ .

${m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$

Rearrange the equation ${m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$ for the velocity ${v_1}$ ,

${v_1} = \frac{{{m_2}{u_2} - {m_2}{v_2}}}{{{m_1}}}$

Substitute ${\rm{15}}{\rm{.0 kg}}$ for ${m_1}$ , ${\rm{3}}{\rm{.00 kg}}$ for ${m_2}$ , $8.00{\rm{ m/s}}$ for ${u_2}$ , and $- 2.00{\rm{ m/s}}$ for ${v_2}$ in equation ${v_1} = \frac{{{m_2}{u_2} - {m_2}{v_2}}}{{{m_1}}}$ .

$\begin{array}{c}\\{v_1} = \frac{{\left( {{\rm{3}}{\rm{.00 kg}}} \right)\left( {{\rm{8}}{\rm{.00 m/s}}} \right) - \left( {{\rm{3}}{\rm{.00 kg}}} \right)\left( {{\rm{ - 2}}{\rm{.00 m/s}}} \right)}}{{{\rm{15}}{\rm{.0 kg}}}}\\\\ = 2.00{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\\end{array}$

Find the maximum distance up to which the spring compresses after the collision.

The kinetic energy of the stone before collision is equal to the energy stored in the spring after the collision.

The kinetic energy ${K_1}$ of the block after collision is,

${K_2} = \frac{1}{2}{m_2}{v_2}^2$

The potential energy U of the system with spring constant k is,

$U = \frac{1}{2}k{x^2}$

From law of conservation of energy,

$U = {K_2}$

Substitute $\frac{1}{2}k{x^2}$ for U and $\frac{1}{2}{m_2}{v_2}^2$ for ${K_2}$ and solve for x.

$\begin{array}{c}\\\frac{1}{2}k{x^2} = \frac{1}{2}{m_2}{v_2}^2\\\\x = \sqrt {\frac{{{m_2}{v_2}^2}}{k}} \\\end{array}$

Substitute ${\rm{15}}{\rm{.0 kg}}$ for ${m_1}$ , ${\rm{400 N/m}}$ for $k$ , and $- 2.00{\rm{ m/s}}$ for ${v_2}$ in equation $x = \sqrt {\frac{{{m_2}{v_2}^2}}{k}}$ .

$\begin{array}{c}\\x = \sqrt {\frac{{\left( {{\rm{15}}{\rm{.0 kg}}} \right){{\left( { - 2.00{\rm{ m/s}}} \right)}^2}}}{{\left( {{\rm{400 N/m}}} \right)}}} \\\\ = 0.39{\rm{ m}}\\\end{array}$

Ans:

The maximum compression in the spring after the collision is $0.39{\rm{ m}}$ .