Chapter 11, Problem 55PQ

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. They collide and stick together, as the driver of one car reaches out and grabs hold of the other driver. The two bumper cars move off together after the collision, and friction is negligible between the cars and the ground.

1. a. If the masses of bumper cars 1 and 2 are 596 kg and 625 kg respectively, what is the velocity of the bumper cars immediately after the collision?
2. b. What is the kinetic energy lost in the collision?
3. c. Compare your answers to part (b) from this and Problem 54. Is one answer larger than the other? Discuss and explain any differences you find.

To determine

The velocity of the bumper cars after collision.

Answer to Problem 55PQ

The velocity of the bumper cars after collision is $\underset{\_}{4.44\text{m/s at 2}\text{.38}°\text{ west of south}}$.

Explanation of Solution

Consider positive $x$ axis points to east, and positive y axis points north. The momentum of two bumper cars is conserved in each dimension.

According to law of conservation of momentum initial momentum in x direction is equal to the final total momentum in x direction.

  $p_{{\text{tot}}_{xi}}=p_{{\text{tot}}_{xf}}$                                                                                                                (I)

Rewrite equation (I) in terms of mass and velocity

  $m_1{v}_{1xi}-m_2{v}_{2xi}=\left(m_1+m_2\right)v_{2xf}$                                                                                  (II)

Here, $m_1$ is the mass of car 1, $m_2$ is the mass of car 2, $v_{1xi}$ is the velocity of car 1, $v_{2xi}$ is the velocity of car 2, and $v_{xf}$ is the final velocity.

Car 1 is moving towards east and car 2 moves towards south of the west. Since two cars are moving in opposite direction $v_{2i}$ will be negative.

Rearrange equation (II) to obtain an expression for $v_{xf}$.

  $v_{2xf}=\frac{m_1{v}_{1xi}-m_2{v}_{2xi}}{\left(m_1+m_2\right)}$                                                                                                (III)

Similarly initial momentum in y direction is equal to the final momentum in y direction.

  $p_{{\text{tot}}_{yi}}=p_{{\text{tot}}_{yf}}$                                                                                                             (IV)

Similar to equation (III) final velocity in y direction can written in which the y component of velocity of car 1, $v_{1yi}$ will be zero.

Hence, expression for final velocity in y direction is.

  $v_{2yf}=\frac{-m_2{v}_{2yi}}{\left(m_1+m_2\right)}$                                                                                                      (V)

The magnitude of final velocity of car 2 is

  $v_{2f}=\sqrt{{\left(v_{2xf}\right)}^2+{\left(v_{2yf}\right)}^2}$                                                                                         (VI)

The expression for direction of final velocity of car 2.

  $\theta ={\tan }^{-1}\left(\frac{v_{2yf}}{v_{2xf}}\right)$                                                                                                     (VII)

Conclusion:

Substitute, $596\text{kg}$ for $m_1$, $5.62\text{m/s}$ for $v_{1xi}$, $625\text{kg}$ for $m_2$, $10.00\cos \left(60°\right)\text{m/s}$ for $v_{2xi}$ in equation (III).

  $v_{2xf}=\frac{\left(596\text{kg}\right)\left(5.62\text{m/s}\right)-\left(625\text{kg}\right)\left(10.00\cos \left(60°\right)\text{m/s}\right)}{\left(596\text{kg}+625\text{kg}\right)}$                                 (VIII)

Substitute, $596\text{kg}$ for $m_1$, $625\text{kg}$ for $m_2$, $10.00\cos \left(60°\right)\text{m/s}$ for $v_{2yi}$ in equation (V).

  $v_{2yf}=\frac{-\left(625\text{kg}\right)\left(10.00\cos \left(60°\right)\text{m/s}\right)}{\left(596\text{kg}+625\text{kg}\right)}$                                                                   (IX)

Substitute, equation (VIII) and (IX) in (VI).

  $\begin{array}{c}{v}_{2f}=\sqrt{{\left(\frac{\left(596\text{kg}\right)\left(5.62\text{m/s}\right)-\left(625\text{kg}\right)\left(10.00\cos \left(60°\right)\text{m/s}\right)}{\left(596\text{kg}+625\text{kg}\right)}\right)}^2+{\left(\frac{-\left(625\text{kg}\right)\left(10.00\cos \left(60°\right)\text{m/s}\right)}{\left(596\text{kg}+625\text{kg}\right)}\right)}^2}\\ =4.44\text{m/s}\end{array}$

Substitute, equation (VIII) and (IX) in (VII).

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\frac{-\left(625\text{kg}\right)\left(10.00\cos \left(60°\right)\text{m/s}\right)}{\left(596\text{kg}+625\text{kg}\right)}}{\frac{\left(596\text{kg}\right)\left(5.62\text{m/s}\right)-\left(625\text{kg}\right)\left(10.00\cos \left(60°\right)\text{m/s}\right)}{\left(596\text{kg}+625\text{kg}\right)}}\right)\\ =\text{2}\text{.38}°\end{array}$

Therefore, the velocity of the bumper cars after collision is $\underset{\_}{4.44\text{m/s at 2}\text{.38}°\text{ west of south}}$.

To determine

The kinetic energy lost in collision.

Answer to Problem 55PQ

The kinetic energy lost in collision is $\underset{\_}{2.86\times {10}^4\text{J}}$.

Explanation of Solution

Write the expression for lost kinetic energy.

  $\begin{array}{c}\Delta K_{lost}=-\Delta K=K_{to{t}_i}-K_{to{t}_f}\\ =\left[\left(\frac{1}{2}{m}_1{v}_{1i}{}^2\right)+\left(\frac{1}{2}{m}_2{v}_{2i}{}^2\right)-\left(\frac{1}{2}\left(m_1+m_2\right){\left(v_{2f}\right)}^2\right)\right]\end{array}$                                       (X)

Conclusion:

Substitute, $596\text{kg}$ for $m_1$, $5.62\text{m/s}$ for $v_{1xi}$, $625\text{kg}$ for $m_2$, $10.00\text{m/s}$ for $v_{2xi}$, and $4.44\text{m/s}$ for $v_{2f}$ in equation (X).

  $\begin{array}{c}\Delta K_{lost}=\left[\left(\frac{1}{2}\left(596\text{kg}\right){\left(5.62\text{m/s}\right)}^2\right)+\left(\frac{1}{2}\left(625\text{kg}\right){\left(10.00\text{m/s}\right)}^2\right)-\left(\frac{1}{2}\left(596\text{kg}+625\text{kg}\right){\left(4.44\text{m/s}\right)}^2\right)\right]\\ =2.86\times {10}^4\text{J}\end{array}$

Therefore, the kinetic energy lost in collision is $\underset{\_}{2.86\times {10}^4\text{J}}$.

To determine

Compare the answers in part (b) of this and problem 54, and explain if there are any differences.

Answer to Problem 55PQ

The kinetic energy lost is greater in this problem, since collision is completely inelastic.

Explanation of Solution

Kinetic energy lost in previous problem is $1.08\times {10}^4\text{J}$ and in this problem is $2.86\times {10}^4\text{J}$. Thus, Kinetic energy lost is greater in this case compared to problem 54.

In an elastic collision there is no loss in kinetic energy, and total energy is conserved. But in an inelastic collision, there will be a fractional loss in kinetic energy. An extreme case of inelastic collision is the one in which colliding objects stick together. In such situations kinetic energy loss will be maximum.

Conclusion:

Since two cars stick together after collision in this case kinetic energy loss will be greater as compared to problem 54.