Chapter 11, Problem 54PQ

Two bumper cars at the county fair are sliding toward one another (Fig. P11.54). Initially, bumper car 1 is traveling to the east at 5.62 m/s, and bumper car 2 is traveling 60.0° south of west at 10.00 m/s. After they collide, bumper car 1 is observed to be traveling to the west with a speed of 3.14 m/s. Friction is negligible between the cars and the ground. a. If the masses of bumper cars 1 and 2 are 596 kg and 625 kg respectively, what is the velocity of bumper car 2 immediately after the collision? b. What is the kinetic energy lost in the collision?

FIGURE P11.54 Problems 54 and 55.

To determine

Velocity of bumper car 2 after collision.

Answer to Problem 54PQ

Velocity of bumper car 2 after collision is $\underset{\_}{\left(3.35\widehat{i}-8.86\widehat{j}\right)\text{m}/\text{s}}$.

Explanation of Solution

Positive x axis points to the east and the positive y axis points to the north. Apply law of conservation of momentum. The momentum of cars before collision must be equal to the momentum of cars after collision.

  $P_{to{t}_{xi}}=P_{to{t}_{xf}}$                                                                                                              (I)

Here, $P_{to{t}_{xi}}$ is the total momentum of system before collision along the x axis and $P_{to{t}_{xf}}$ is the total momentum after collision along the x axis.

Elaborate equation (I) in terms of mass and velocity.

  $m_1{v}_{1_{xi}}-m_2{v}_{2_{xi}}\cos \theta =-m_1{v}_{1_{xf}}+m_2{v}_{2_{xf}}$                                                                      (II)

Here, $m_1$ is the mass of car 1, $v_{1_{xi}}$ is the velocity of car 1 before collision, $m_2$ is the mass of car 2, $v_{2_{xi}}$ is the velocity of car 2 before collision, $\theta$ is the angle made by car 2 with x axis, $v_{1_{xf}}$ is the velocity of car 1 after collision and $v_{2_{xf}}$ is the velocity of car 2 after collision.

Apply the same condition of conservation of momentum in the y direction also. Here only car 2 travels in the y direction.

The initial total momentum in the y direction is equal to the final momentum in the y direction.

  $P_{to{t}_{yi}}=P_{to{t}_{yf}}$                                                                                                         (III)

Here, $P_{to{t}_{yi}}$ is the total momentum of car before collision in y direction and $P_{to{t}_{yf}}$ is the total momentum of car after collision in the y direction.

Elaborate equation (III) in terms of mass and velocity.

  $-m_2{v}_{2_{yi}}\sin \theta =m_2{v}_{2_{yf}}$                                                                                    (IV)

Here, $m_2$ is the mass of car 2, $v_{2_{y_i}}$ is the velocity of car 2 in the y direction before collision, $\theta$ is the angle made by car 2 with x axis and $v_{2_{yf}}$ is the velocity of car 2 after collision along y axis.

Write the equation to find the final velocity of second car.

  $\overrightarrow{v_f}=\overrightarrow{v_{2_{xf}}}\widehat{i}+\overrightarrow{v_{2_{yf}}}\widehat{j}$                                                                                                     (V)

Here, $\overrightarrow{v_f}$ is the final velocity of car 2 after collision, $\widehat{i}$ and $\widehat{j}$ are unit vectors along x and y directions.

Conclusion:

Substitute $596\text{kg}$ for $m_1$, $5.62\text{m}/\text{s}$ for $v_{1_{xi}}$, $625\text{kg}$ for $m_2$, $10.00\text{m}/\text{s}$ for $v_{2_{xi}}$, $60°$ for $\theta$ and $3.14\text{m}/\text{s}$ for $v_{1_{xf}}$ and simplify for $v_{2_{xf}}$.

  $\begin{array}{c}\left(596\text{kg}\right)\left(5.62\text{m}/\text{s}\right)-\left(625\text{kg}\right)\left[10.00\cos \left(60°\right)\text{m}/\text{s}\right]=-\left(596\text{kg}\right)\left(3.14\text{m}/\text{s}\right)+\left(625\text{kg}\right)v_{2_{xf}}\\ v_{2_{xf}}=3.35\text{m}/\text{s}\end{array}$

Substitute $625\text{kg}$ for $m_2$, $10.00\text{m}/\text{s}$ for $v_{2_{yi}}$ and $60°$ for $\theta$ in equation (IV) and solve for $v_{2_{yf}}$.

  $\begin{array}{c}-\left(625\text{kg}\right)\left[10.00\sin \left(60°\right)\text{m}/\text{s}\right]=\left(625\text{kg}\right)v_{2_{yf}}\\ v_{2_{yf}}=-8.66\text{m}/\text{s}\end{array}$

Substitute $3.35\text{m}/\text{s}$ for $v_{2_{xf}}$ and $-8.86\text{m}/\text{s}$ for $v_{2_{yf}}$ in equation (V) to get $\overrightarrow{v_f}$.

  $\overrightarrow{v_f}=\left(3.35\widehat{i}-8.86\widehat{j}\right)\text{m}/\text{s}$

Therefore, velocity of bumper car 2 after collision is $\underset{\_}{\left(3.35\widehat{i}-8.86\widehat{j}\right)\text{m}/\text{s}}$.

To determine

Kinetic energy lost during the collision.

Answer to Problem 54PQ

The kinetic energy lost is $\underset{\_}{1.08\times {10}^4\text{J}}$.

Explanation of Solution

Write the equation to find the resultant final speed of car 2 after collision.

  $v_{2_f}=\sqrt{v_{2_{xf}}{}^2+v_{2_{yf}}{}^2}$                                                                                                   (VI)

Here, $v_{2_f}$ is the final speed of car 2 after collision.

The kinetic energy lost is equal to the difference between the kinetic energy before collision and after collision.

Write the equation to find the kinetic energy lost.

  $\begin{array}{c}{K}_{lost}=K_{to{t}_i}-K_{to{t}_f}\\ =\left[\frac{1}{2}{m}_1{v}_{1_{xi}}{}^2+\frac{1}{2}{m}_2{v}_{2_{xi}}{}^2\right]-\left[\frac{1}{2}{m}_1{v}_{1_{yf}}{}^2+\frac{1}{2}{m}_2{v}_{2_f}{}^2\right]\end{array}$                                            (VII)

Here, $K_{lost}$ is the kinetic energy lost, $K_{to{t}_i}$ is the total kinetic energy before collision and $K_{to{t}_f}$ is the kinetic energy after collision.

Conclusion:

Substitute $3.35\text{m}/\text{s}$ for $v_{2_{xf}}$ and $-8.66\text{m}/\text{s}$ for $v_{2_{yf}}$ in equation (VI) to get $v_{2_f}$.

  $\begin{array}{c}{v}_{2_f}=\sqrt{{\left(3.35\text{m}/\text{s}\right)}^2+{\left(-8.66\text{m}/\text{s}\right)}^2}\\ =9.28\text{m}/\text{s}\end{array}$

Substitute $596\text{kg}$ for $m_1$, $5.62\text{m}/\text{s}$ for $v_{1_{xi}}$, $625\text{kg}$ for $m_2$, $10.00\text{m}/\text{s}$ for $v_{2_{xi}}$, $3.14\text{m}/\text{s}$ for $v_{1_{yf}}$ and $9.28\text{m}/\text{s}$ for $v_{2_f}$ in equation (VII) to get $K_{lost}$.

$\begin{array}{c}{K}_{lost}=\left[\frac{1}{2}\left(596\text{kg}\right){\left(5.62\text{m}/\text{s}\right)}^2+\frac{1}{2}\left(625\text{kg}\right){\left(10.0\text{m}/\text{s}\right)}^2\right]-\left[\frac{1}{2}\left(596\text{kg}\right){\left(3.14\text{m}/\text{s}\right)}^2+\frac{1}{2}\left(625\text{kg}\right){\left(9.28\text{m}/\text{s}\right)}^2\right]\\ =1.08\times {10}^4\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\left(\frac{\text{1J}}{1\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =1.08\times {10}^4\text{J}\end{array}$

Therefore, the kinetic energy lost is $\underset{\_}{1.08\times {10}^8\text{J}}$.