Question

Block 1, of mass $m_1$ = 3.70 , moves along a frictionless air track with speed $v_1$ = 23.0 . It collides with block 2, of mass $m_2$ = 13.0 , which was initially at rest. The blocks stick togetherafter the collision.

$MLM_in_5.jpg$

part A

Find the magnitude $p_i$ of the total initial momentum of the two-blocksystem.

Express your answernumerically.

part B

Find $v_f$ , the magnitude of the final velocity of the two-blocksystem.

Express your answer numerically

part C

What is the change $\Delta K = K_{\rm final} - K_{\rm initial}$ in the system's kinetic energy due to thecollision?

Express your answer numerically injoules.

Answer 1

Concepts and reason

The concept required to solve this problem is Conservation of momentum.

Initially, calculate the momentum of each block before the collision and add them to get the total initial momentum.

Later, use the conservation of momentum to solve for final velocity of the two-block system.

Finally, use the kinetic energy formula to solve for total initial and final kinetic energies and calculate the difference between the two.

Fundamentals

The momentum of an object is the product of its mass and velocity. The magnitude of momentum is expressed as,

$p = mv$

Here, $p$ is the momentum, $m$ is the mass and $v$ is the speed.

The conservation of momentum principle says that total momentum of an isolated system is always conserved. This can be expressed as,

${p_{{\rm{initial}}}} = {p_{{\rm{final}}}}$

Here, ${p_{{\rm{initial}}}}$ is the total initial momentum of the system and ${p_{{\rm{final}}}}$ is the total final momentum.

The kinetic energy is given as,

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

Here, $m$ is the mass of the object, and $v$ is the speed.

(A)

Calculate the momentum as follows:

Substitute ${p_{1{\rm{i}}}}$ for $p$, $3.70{\rm{ kg}}$ for $m$, and $23.0{\rm{ m/s}}$ for $v$ in the equation $p = mv$ to calculate the initial momentum of the block 1, ${p_{1{\rm{i}}}}$.

$\begin{array}{c}\\{p_{1{\rm{i}}}} = \left( {3.70{\rm{ kg}}} \right)\left( {23.0{\rm{ m/s}}} \right)\\\\ = 85.1{\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}$

Substitute ${p_{{\rm{2i}}}}$ for $p$, ${\rm{13}}{\rm{.0 kg}}$ for $m$, and $0{\rm{ m/s}}$ for $v$ in the equation $p = mv$ to calculate the initial momentum of the block 2, ${p_{{\rm{2i}}}}$.

$\begin{array}{c}\\{p_{{\rm{2i}}}} = \left( {{\rm{13}}{\rm{.0 kg}}} \right)\left( {0{\rm{ m/s}}} \right)\\\\ = 0\\\end{array}$

The total initial momentum is the sum of initial momentum of block 1 and block 2.

${p_{\rm{i}}} = {p_{{\rm{1i}}}} + {p_{{\rm{2i}}}}$

Substitute $85.1{\rm{ kg}} \cdot {\rm{m/s}}$ for ${p_{{\rm{1i}}}}$ and $0$ for ${p_{{\rm{2i}}}}$ in the equation ${p_{\rm{i}}} = {p_{{\rm{1i}}}} + {p_{{\rm{2i}}}}$.

$\begin{array}{c}\\{p_{\rm{i}}} = 85.1{\rm{ kg}} \cdot {\rm{m/s}} + 0\\\\ = 85.1{\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}$

(B)

The total final momentum is the sum of final momentum of block 1 and 2.

${p_{{\rm{final}}}} = {p_{1{\rm{f}}}} + {p_{2{\rm{f}}}}$

Here, ${p_{1{\rm{f}}}}$ is the final momentum of the block 1 and ${p_{{\rm{2f}}}}$ is the final momentum of block 2.

Substitute ${m_1}{v_{\rm{f}}}$ for ${p_{{\rm{1f}}}}$ and ${m_2}{v_{\rm{f}}}$ for ${p_{2{\rm{f}}}}$ in the above equation ${p_{{\rm{final}}}} = {p_{1{\rm{f}}}} + {p_{2{\rm{f}}}}$.

${p_{{\rm{final}}}} = {m_1}{v_{\rm{f}}} + {m_2}{v_{\rm{f}}}$

Substitute ${m_1}{v_{\rm{f}}} + {m_2}{v_{\rm{f}}}$ for ${p_{{\rm{final}}}}$ in the conservation of momentum equation ${p_{{\rm{initial}}}} = {p_{{\rm{final}}}}$ and solve for ${v_{\rm{f}}}$.

$\begin{array}{c}\\{p_{{\rm{initial}}}} = {m_1}{v_{\rm{f}}} + {m_2}{v_{\rm{f}}}\\\\{v_{\rm{f}}} = \frac{{{p_{{\rm{initial}}}}}}{{{m_1} + {m_2}}}\\\end{array}$

Substitute $85.1{\rm{ kg}} \cdot {\rm{m/s}}$ for ${p_{{\rm{initial}}}}$, $3.70{\rm{ kg}}$ for ${m_1}$, and $13.0{\rm{ kg}}$ for ${m_2}$ in the equation ${v_{\rm{f}}} = \frac{{{p_{{\rm{initial}}}}}}{{{m_1} + {m_2}}}$ and calculate ${v_{\rm{f}}}$.

$\begin{array}{c}\\{v_{\rm{f}}} = \frac{{85.1{\rm{ kg}} \cdot {\rm{m/s}}}}{{3.70{\rm{ kg}} + 13.0{\rm{ kg}}}}\\\\ = 5.1{\rm{ m/s}}\\\end{array}$

Part C

Use the kinetic energy equation.

Substitute ${K_{1{\rm{i}}}}$ for $K$, ${m_1}$ for $m$, and ${v_1}$ for $v$ in the kinetic energy equation $K = \frac{1}{2}m{v^2}$.

${K_{{\rm{1i}}}} = \frac{1}{2}{m_1}v_1^2$

Here, ${K_{1{\rm{i}}}}$ is the initial kinetic energy of the block 1, ${m_1}$ is the mass of block 1 and, ${v_1}$ is the velocity of block 1.

Substitute ${K_{1{\rm{f}}}}$ for $K$, ${m_1}$ for $m$, and ${v_{\rm{f}}}$ for $v$ in the kinetic energy equation $K = \frac{1}{2}m{v^2}$.

${K_{{\rm{1f}}}} = \frac{1}{2}{m_1}v_{\rm{f}}^2$

Here, ${K_{1{\rm{f}}}}$ is the final kinetic energy of the block 1, ${m_1}$ is the mass of block 1 and, ${v_{\rm{f}}}$ is the final velocity of block 1.

Substitute ${K_{{\rm{2i}}}}$ for $K$, ${m_2}$ for $m$, and ${v_{\rm{f}}}$ for $v$ in the kinetic energy equation $K = \frac{1}{2}m{v^2}$.

${K_{{\rm{2f}}}} = \frac{1}{2}{m_2}v_{\rm{f}}^2$

Here, ${K_{{\rm{2f}}}}$ is the final kinetic energy of the block 2, ${m_2}$ is the mass of block 2 and, ${v_{\rm{f}}}$ is the velocity of block 2.

The total initial kinetic energy is the sum of initial kinetic energy of block 1 and 2.

${K_{{\rm{initial}}}} = {K_{1{\rm{i}}}} + {K_{2{\rm{i}}}}$

Here, ${K_{{\rm{2i}}}}$ is the initial kinetic energy of the block 2.

Substitute $\frac{1}{2}{m_1}v_1^2$ for ${K_{1{\rm{i}}}}$ and $0$ for ${K_{2{\rm{i}}}}$ in the above equation ${K_{{\rm{initial}}}} = {K_{1{\rm{i}}}} + {K_{2{\rm{i}}}}$.

$\begin{array}{c}\\{K_{{\rm{initial}}}} = \frac{1}{2}{m_1}v_1^2 + 0\\\\ = \frac{1}{2}{m_1}v_1^2\\\end{array}$

The total final kinetic energy is the sum of final kinetic energy of block 1 and 2.

${K_{{\rm{final}}}} = {K_{1{\rm{f}}}} + {K_{2{\rm{f}}}}$

Substitute $\frac{1}{2}{m_1}v_{\rm{f}}^2$ for ${K_{1{\rm{f}}}}$ and $\frac{1}{2}{m_2}v_{\rm{f}}^2$ for ${K_{{\rm{2f}}}}$ in the above equation ${K_{{\rm{final}}}} = {K_{1{\rm{f}}}} + {K_{2{\rm{f}}}}$.

${K_{{\rm{final}}}} = \frac{1}{2}{m_1}v_{\rm{f}}^2 + \frac{1}{2}{m_2}v_{\rm{f}}^2$

The change in the kinetic energy due to collision is,

$\Delta K = {K_{{\rm{final}}}} - {K_{{\rm{initial}}}}$

Substitute $\frac{1}{2}{m_1}v_{\rm{f}}^2 + \frac{1}{2}{m_{\rm{2}}}v_{\rm{f}}^2$ for ${K_{{\rm{final}}}}$ and $\frac{1}{2}{m_1}v_1^2$ for ${K_{{\rm{initial}}}}$ in the above equation $\Delta K = {K_{{\rm{final}}}} - {K_{{\rm{initial}}}}$.

$\Delta K = \frac{1}{2}{m_1}v_{\rm{f}}^2 + \frac{1}{2}{m_{\rm{2}}}v_{\rm{f}}^2 - \frac{1}{2}{m_1}v_1^2$

Substitute $3.70{\rm{ kg}}$ for ${m_1}$, $13.0{\rm{ kg}}$ for ${m_2}$, $5.1{\rm{ m/s}}$ for ${v_{\rm{f}}}$, and $23.0{\rm{ m/s}}$ for ${v_1}$ in the above equation $\Delta K = \frac{1}{2}{m_1}v_{\rm{f}}^2 + \frac{1}{2}{m_{\rm{2}}}v_{\rm{f}}^2 - \frac{1}{2}{m_1}v_1^2$ and calculate $\Delta K$.

$\begin{array}{c}\\\Delta K = \frac{1}{2}\left( {3.70{\rm{ kg}}} \right){\left( {5.1{\rm{ m/s}}} \right)^2} + \frac{1}{2}\left( {13.0{\rm{ kg}}} \right)\left( {5.1{\rm{ m/s}}} \right) - \frac{1}{2}\left( {3.70{\rm{ kg}}} \right){\left( {23.0{\rm{ m/s}}} \right)^2}\\\\ = - 761.82{\rm{ J}}\\\end{array}$

Ans: Part A

The magnitude of total initial momentum of the two-block system is $85.1{\rm{ kg}} \cdot {\rm{m/s}}$.

Part B

The magnitude of the final velocity of the two-block system is 5.1 m/s.

Part C

The change in the kinetic energy due to collision is $- 761.82{\rm{ J}}$.