Chapter 9, Problem 35P

(a)

To determine

Velocity of $0.300\text{kg}$ puck after the collision.

Answer to Problem 35P

Velocity of $0.300\text{kg}$ puck after the collision is $1.07\text{m/s}$.

Explanation of Solution

Position of the puck before collision.

After the collision of two puck, resolve the component of both the puck in x-y direction.

Write the expression for the conservation of momentum in x direction.

    $m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}\cos \varphi +m_2{v}_{2f}\cos \theta$

Here, $m_1$ is the mass of the puck at rest, $v_{1i}$  is the initial velocity of the puck of mass $m_1$, $v_{1f}$ is the final velocity of the puck which was initially at rest, $v_{2i}$ is the initial velocity of puck of mass $m_2$, $v_{2f}$ is the final velocity of  puck of mass $m_2$ , $\theta$  is the angle made with horizontal by puck of mass $m_2$ and $\varphi$ is the angle made with horizontal by puck of mass $m_1$.

Substitute $0\text{m/s}$ for $v_{1i}$ and rearrange above equation for value of $v_{1f}\cos \varphi$.

    $v_{1f}\cos \varphi =\frac{\left(m_2{v}_{2i}-m_2{v}_{2f}\cos \theta \right)}{m_1}$                                                                    (I)

Initially velocity in y direction is zero.

Write the expression for conservation of momentum in y-direction.

  $0=m_1{v}_{1f}\sin \varphi -m_2{v}_{2f}\sin \theta$

Solve above equation for value of $v_{1f}\sin \varphi$.

    $v_{1f}\sin \varphi =\frac{\left(m_2{v}_{2f}\sin \theta \right)}{m_1}$                                                                               (II)

Divide equation (II) by equation (I) for the value of $\varphi$.

    $\tan \varphi =\frac{\left(m_2{v}_{2f}\sin \theta \right)}{\left(m_2{v}_{2i}-m_2{v}_{2f}\cos \theta \right)}$

Solve the above equation for $\varphi$ .

    $\varphi ={\tan }^{-1}\left(\frac{\left(m_2{v}_{2f}\sin \theta \right)}{\left(m_2{v}_{2i}-m_2{v}_{2f}\cos \theta \right)}\right)$                                                              (III)

Conclusion:

Substitute $0.200\text{kg}$ for $m_2$, $1\text{m/s}$ for $v_{2f}$ , $53°$ for $\theta$ and $0.300\text{kg}$ for $m_1$ and $2\text{m/s}$ for $v_{2i}$ in equation (III) for value of $\varphi$ .

  $\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{\left(\left(0.200kg\right)\left(1\text{m/s}\right)\sin \left(53°\right)\right)}{\left(\left(0.200\text{kg}\right)\left(2\text{m/s}\right)-\left(0.200\text{kg}\right)\left(1\text{m/s}\right)\cos \left(53°\right)\right)}\right)\\ =29.73°\end{array}$

Substitute $0.200\text{kg}$ for $m_2$, $1\text{m/s}$ for $v_{2f}$ , $53°$ for $\theta$, $29.73°$ for $\varphi$ in equation (II) and solve for $v_{1f}$ .

  $\begin{array}{c}{v}_{1f}=\frac{\left(0.200\text{kg}\right)\left(1\text{m/s}\right)\sin \left(53°\right)}{\left(0.300\text{kg}\right)\left(\sin \left(29.73°\right)\right)}\\ =1.073\text{m/s}\end{array}$

Thus, velocity of $0.300\text{kg}$ puck after the collision is $1.07\text{m/s}$.

(b)

To determine

Fraction of kinetic energy transferred away or transformed to other forms of energy in collision.

Answer to Problem 35P

Fraction of kinetic energy transferred away or transformed to other forms of energy in collision is $-0.32$.

Explanation of Solution

To calculate fraction of kinetic energy , first calculate the change in kinetic energy.

Write the expression for the kinetic energy before collison.

    $K_i=\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{2i}^2$                                                                                  (IV)

Here, $K_i$ is the initial kinetic energy.

Write the expression for the final kinetic energy.

    $K_f=\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{v}_{2f}^2$                                                                                 (V)

Here, $K_f$  is the final kinetic energy after collision.

Write the expression for the change in kinetic energy.

  $\Delta K=K_f-K_i$                                                                                              (VI)

Substitute $\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{2i}^2$ for $K_i$ and $\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{v}_{2f}^2$ for $K_f$ in equation (VI) for $\Delta K$.

    $\Delta K=\left(\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{v}_{2f}^2\right)-\left(\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{2i}^2\right)$                                       (VII)

Fraction of kinetic energy is the ratio of change in kinetic energy to the initial kinetic energy.

    $f=\frac{\Delta K}{K_i}$                                                                                                   (VIII)

Here, $f$ is the ratio of change in kinetic energy to the initial kinetic energy.

Substitute $\left(\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{v}_{2f}^2\right)-\left(\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{2i}^2\right)$ for $\Delta K$ and $\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{2i}^2$ for $K_i$ in equation (VIII).

    $f=\frac{\left(\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{v}_{2f}^2\right)-\left(\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{2i}^2\right)}{\left(\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{2i}^2\right)}$                                            (IX)

Conclusion:

Substitute $0.300\text{kg}$ for $m_1$, $1.07\text{m/s}$ for $v_{1f}$, $0.300\text{kg}$ for $m_2$ , $1\text{m/s}$ for $v_{2f}$ and $0\text{m/s}$ for $v_{1i}$ and $2\text{m/s}$ for $v_{2i}$ in equation (IX).

    $\begin{array}{c}f=\frac{\left[\begin{array}{l}\left(\frac{\text{1}}{\text{2}}\text{(0}\text{.300kg)}{\left(\text{1}\text{.073}\right)}^{\text{2}}\text{+}\frac{\text{1}}{\text{2}}\left(\text{0}\text{.200kg}\right){\left(\text{1m/s}\right)}^{\text{2}}\right)\\ \text{-}\left(\frac{\text{1}}{\text{2}}\left(\text{0}\text{.300kg}\right){\left(\text{0m/s}\right)}^{\text{2}}\text{+}\frac{\text{1}}{\text{2}}\left(\text{0}\text{.200kg}\right){\left(\text{2m/s}\right)}^{\text{2}}\right)\end{array}\right]}{\left(\frac{\text{1}}{\text{2}}\left(\text{0}\text{.300kg}\right){\left(\text{0m/s}\right)}^{\text{2}}\text{+}\frac{\text{1}}{\text{2}}\left(\text{0}\text{.200kg}\right){\left(\text{2m/s}\right)}^{\text{2}}\right)}\\ =\frac{-\left(0.2546\right)}{\left(0.8\right)}\\ =-0.32\end{array}$

Thus, fraction of kinetic energy transferred away or transformed to other forms of energy in collision is $-0.32$.