Chapter 7, Problem 23P

To determine

The speed and direction of each object after the collision.

Answer to Problem 23P

The speed of ice puck after the collision $v_1=-1\text{m}/\text{s}$ towards west direction.

The speed of puck after the collision $v_2=2\text{m}/\text{s}$ towards east direction.

Explanation of Solution

Given:

The mass of ice puck is $m_1=0.450\text{ kg}$ .

The initial velocity of the ice puck is $u_1=3.0\text{m}/\text{s}$ .

The mass of puck is $m_2=0.900\text{ kg}$ .

The initial velocity of the puck is $u_2=0$ .

Formula used:

The expression for law of 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 ice puck, $m_2$ is the mass of puck, $u_1$ is the initial velocity of ice puck, $u_2$ is the initial velocity of the puck, $v_1$ is the final velocity of ice puck, and $v_2$ is the final velocity of the puck.

The expression for conservation of kinetic energy is,

  $\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$

Here, $m_1$ is the mass of the ice puck, $m_2$ is the mass of puck, $u_1$ is the initial velocity of ice puck,

  $u_2$ is the initial velocity of the puck, $v_1$ is the final velocity of ice puck, and $v_2$ is the final velocity of the puck.

Calculation:

The speed of each object using the relation of conservation of momentum is,

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\\ \left[\begin{array}{l}\left(0.450\text{ kg}\right)\left(3.0\text{m}/\text{s}\right)\\ +\left(0.900\text{ kg}\right)\left(0\right)\end{array}\right]=\left(0.450\text{ kg}\right)v_1+\left(0.900\text{ kg}\right)v_2\\ 1.35=0.45v_1+0.90v_2\\ 3=v_1+2v_2\end{array}$

  $v_1=3-2v_2$  .........(1)

Given that the collision is perfectly elastic. The speed of each object using the relation of conservation of kinetic energy is,

  $\begin{array}{l}\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2\\ \left[\begin{array}{l}\frac{1}{2}\left(0.450\text{ kg}\right){\left(3.0\text{m}/\text{s}\right)}^2\\ +\frac{1}{2}\left(0.900\text{ kg}\right){\left(0\right)}^2\end{array}\right]=\frac{1}{2}\left(0.450\text{ kg}\right)v_1^2+\frac{1}{2}\left(0.900\text{ kg}\right)v_2^2\\ 4.05=0.45v_1^2+0.90v_2^2\\ 9=v_1^2+2v_2^2\end{array}$

  $\begin{array}{l}9={\left(3-2v_2\right)}^2+2v_2^2\\ 9=9+4v_2^2-12v_2+2v_2^2\\ 6v_2^2=12v_2\\ v_2=2\text{m}/\text{s}\end{array}$

Hence, the speed of puck after the collision $v_2=2\text{m}/\text{s}$ towards east direction.

The speed of ice puck after the collision using the relation is,

  $\begin{array}{l}{v}_1=3-2v_2\\ v_1=3-2\times \left(2\text{m}/\text{s}\right)\\ v_1=-1\text{m}/\text{s}\end{array}$

Hence, the speed of ice puck after the collision $v_1=-1\text{m}/\text{s}$ towards west direction.

Conclusion:

Thus, the speed of ice puck after the collision $v_1=-1\text{m}/\text{s}$ towards west direction.

Thus, the speed of puck after the collision $v_2=2\text{m}/\text{s}$ towards east direction.