Question

On a frictionless, horizontal air table, puck A (with mass 0.250 kg ) is moving toward puck B (with mass 0.360 kg ), that is initially at rest. After the collision, puck A has a velocity of 0.125 m/s to the left, and puck B has velocity 0.660 m/s to the right.



A- What was the speed of puck A before the collision?


B- Calculate the change in the total kinetic energy of the system that occurs during the collision.

Answer 1

Concepts and reason

The concepts required to solve the given problem are law of conservation of kinetic energy and liner momentum.

Initially, calculate the speed of the puck A before the collision using law of conservation of momentum. Later, calculate the initial and final kinetic energy of the system that occurs during the collision. Finally, calculate the change in kinetic energy.

Fundamentals

The expression of the kinetic energy in terms of momentum is,

$KE = \frac{{{p^2}}}{{2m}}$

Here, KE is the kinetic energy, p is the momentum, and m is the mass.

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,

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

Here, $K{E_{\rm{i}}}$ is the initial kinetic energy and $K{E_{\rm{f}}}$ is the final kinetic energy.

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.

(A)

According to the law of conservation of momentum,

${m_{\rm{A}}}{\vec u_{\rm{A}}} + {m_{\rm{B}}}{\vec u_{\rm{B}}} = {m_{\rm{A}}}{\vec v_{\rm{A}}} + {m_{\rm{B}}}{\vec v_{\rm{B}}}$ …… (1)

Here, ${m_{\rm{A}}}$ and ${m_{\rm{B}}}$ are the masses of pucks A and B respectively, ${\vec u_{\rm{A}}}$ and ${\vec u_{\rm{B}}}$ are the initial speeds of pucks A and B before collision, ${\vec v_{\rm{A}}}$ and ${\vec v_{\rm{B}}}$ are final speed after collision.

Rearrange equation (1) for ${u_{\rm{A}}}$ .

${\vec u_{\rm{A}}} = \frac{{{m_{\rm{A}}}{{\vec v}_{\rm{A}}} + {m_{\rm{B}}}{{\vec v}_{\rm{B}}} - {m_{\rm{B}}}{{\vec u}_{\rm{B}}}}}{{{m_{\rm{A}}}}}$

Substitute 0.250 kg for ${m_{\rm{A}}}$ , 0.360 kg for ${m_{\rm{B}}}$ , $\left( { - 0.125{\rm{ m/s}}} \right)$ for ${\vec v_{\rm{A}}}$ , 0 m/s for ${\vec u_{\rm{B}}}$ , and 0.660 m/s for ${\vec v_{\rm{B}}}$ .

$\begin{array}{c}\\{u_{\rm{A}}} = \left( {\frac{{\left( {0.250{\rm{ kg}}} \right)\left( { - 0.125{\rm{ m/s}}} \right){\rm{ + }}\left( {0.360{\rm{ kg}}} \right)\left( {0.660{\rm{ m/s}}} \right) - \left( {0.360{\rm{ kg}}} \right)\left( {0{\rm{ m/s}}} \right){\rm{ }}}}{{\left( {0.250{\rm{ kg}}} \right)}}} \right)\\\\ = \left( {\frac{{\left( { - 0.03125{\rm{ kg}} \cdot {\rm{m/s}}} \right){\rm{ + }}\left( {0.2376{\rm{ kg}} \cdot {\rm{m/s}}} \right){\rm{ }}}}{{\left( {0.250{\rm{ kg}}} \right)}}} \right)\\\\ = 0.825{\rm{ m/s}}\\\end{array}$

(B)

The initial kinetic energy before collision is given by the following formula.

${K_i} = \frac{1}{2}{m_{\rm{A}}}u_{\rm{A}}^2 + \frac{1}{2}{m_{\rm{B}}}u_{\rm{B}}^2$

Here, ${K_i}$ is the initial kinetic energy, ${m_{\rm{A}}}$ and ${m_{\rm{B}}}$ are the masses of pucks A and B respectively, ${u_{\rm{A}}}$ and ${u_{\rm{B}}}$ are the initial speeds of pucks A and B before collision.

Substitute 0.250 kg for ${m_{\rm{A}}}$ , 0.825 m/s for ${u_A}$ , 0.360 kg for ${m_{\rm{B}}}$ , and 0 m/s for ${u_B}$ in the above equation.

$\begin{array}{c}\\{K_{\rm{i}}} = \frac{1}{2}{m_{\rm{A}}}u_{\rm{A}}^2 + \frac{1}{2}{m_{\rm{B}}}u_{\rm{B}}^2\\\\ = \frac{1}{2}\left( {0.250{\rm{ kg}}} \right){\left( {0.8254{\rm{ m/s}}} \right)^2} + \frac{1}{2}\left( {0.360{\rm{ kg}}} \right){\left( {0{\rm{ m/s}}} \right)^2}\\\\ = 0.08516{\rm{ J}}\\\end{array}$

The final kinetic energy after the collision is given by the following expression:

${K_{\rm{f}}} = \frac{1}{2}{m_{\rm{A}}}v_{\rm{A}}^2 + \frac{1}{2}{m_{\rm{B}}}v_{\rm{B}}^2$

Here, ${K_{\rm{f}}}$ is the final kinetic energy, ${m_{\rm{A}}}$ and ${m_{\rm{B}}}$ are the masses of pucks A and B respectively ${\vec v_{\rm{A}}}$ and ${\vec v_{\rm{B}}}$ are final speeds of pucks A and B after collision.

Substitute 0.250 kg for ${m_{\rm{A}}}$ , $- 0.125{\rm{ m/s}}$ for ${v_{\rm{A}}}$ , 0.360 kg for ${m_{\rm{B}}}$ , and 0.660 m/s for ${v_{\rm{B}}}$ in the above equation.

$\begin{array}{c}\\{K_{\rm{f}}} = \frac{1}{2}\left[ {\left( {0.250{\rm{ kg}}} \right){{\left( { - 0.125{\rm{ m/s}}} \right)}^2} + \left( {0.360{\rm{ kg}}} \right){{\left( {0.660{\rm{ m/s}}} \right)}^2}} \right]\\\\ = 0.080361{\rm{ J}}\\\end{array}$

The change in kinetic energy is given by the following expression:

$\Delta K = {K_{\rm{f}}} - {K_{\rm{i}}}$

Here, ${K_{\rm{f}}}$ and ${K_{\rm{i}}}$ are the final and initial kinetic energies.

Substitute 0.080361 J for ${K_{\rm{f}}}$ and 0.08516 J for ${K_{\rm{i}}}$ .

$\begin{array}{c}\\\Delta K = {K_{\rm{f}}} - {K_{\rm{i}}}\\\\ = 0.080361{\rm{ J}} - {\rm{0}}{\rm{.08516 J}}\\\\{\rm{ = }} - 0.004799{\rm{ J}}\\\end{array}$

Here, the negative sign shows the loss in kinetic energy.

Ans: Part A

The speed of the puck A before the collision was 0.825 m/s.