Chapter 11, Problem 49PQ

Two skateboarders, with masses m₁ = 75.0 kg and m₂ = 65.0 kg, simultaneously leave the opposite sides of a frictionless half-pipe at height h = 4.00 m as shown in Figure P11.49. Assume the skateboarders undergo a completely elastic head-on collision on the horizontal segment of the half-pipe. Treating the skateboarders as particles and assuming they don’t fall off their skateboards, what is the height reached by each skateboarder after the collision?

FIGURE P11.49

To determine

The height reached by each skateboarder after the collision.

Answer to Problem 49PQ

The height reached by skateboarder of mass $75.0\text{kg}$ after the collision is $2.94\text{m}$ and the height reached by skateboarder of mass $65.0\text{kg}$ after the collision is $5.22\text{m}$ .

Explanation of Solution

Apply conservation of mechanical energy for each skateboarder as they slide down the surface and collide.

  $K_i+U_{gi}=K_f+U_{gf}$                                                                                                  (I)

Here, $K_i$ is the kinetic energy of the skateboarder at the maximum height, $U_{gi}$ is the potential energy of the skateboarder at the maximum height, $K_f$ is the kinetic energy of the skateboarder just before collision and $U_{gf}$ is the potential energy of the skateboarder at the maximum height.

The collision occur at $h=0$, therefore the final potential energy is zero. Let $v$ be the speed of skateboarder just before collision. Since they started from rest, initial height kinetic energy is zero.

Write the expression for $K_i$ .

  $K_i=0\text{J}$                                                                                                                 (II)

Write the expression for $U_{gi}$ .

  $U_{gi}=mgh$                                                                                                              (III)

Here, $m$ is the mass of skateboarder,$g$ is the acceleration due to gravity and $h$ is the height of the half pipe.

Write the expression for $K_f$ .

  $K_f=\frac{1}{2}m{v}^2$                                                                                                           (IV)

Here, and $v$ is the speed of skateboarder just before collision.

Write the expression for $U_{gf}$ .

  $U_{gf}=0\text{J}$                                                                                                               (V)

Substitute equations (II),(III),(IV) and (V) in equation (I) to get expression of $v$ .

  $\begin{array}{c}0\text{J}+mgh=\frac{1}{2}m{v}^2+0\text{J}\\ \frac{1}{2}{v}^2=gh\\ v=\sqrt{2gh}\end{array}$                                                                                           (VI)

Rearrange above equation to get $h$ .

  $h=\frac{v^2}{2g}$                                                                                                                  (VII)

Equation (VI) indicates that each skateboarder has the same speed when he meets the other skateboarder since they started from the same height.

Let $v_{1i}$ be the speed of first skateboarder with mass $m_1=75.0\text{kg}$ and $v_{2i}$ is the speed of second skater with mass $m_1=75.0\text{kg}$ just before collision.

Since both are at same speed but in opposite direction, $v_{1i}=-v_{2i}=v$ .

Substitute $9.81{\text{m}/\text{s}}^2$ for $g$ and $4.00\text{m}$ for $h$ in equation (VI) to get $v$ .

  $\begin{array}{c}v=\sqrt{2\left(9.81{\text{m}/\text{s}}^2\right)\left(4.00\text{m}\right)}\\ =8.86\text{m}/\text{s}\end{array}$

Thus, $v_{1i}=-v_{2i}=8.86\text{m}/\text{s}$ .

Write the expression for the velocity of skater with mass $m_1$  after collision.

  $v_{1f}=\left(\frac{m_1-m_2}{m_1+m_2}\right)v_{1i}+\left(\frac{2m_2}{m_1+m_2}\right)v_{2i}$                                                                      (VIII)

Here,$v_{1f}$ is the velocity of the first skater after collision, $m_1$ is the mass of first skater and $m_2$ is the mass of second skater.

Write the expression for the velocity of skater with mass $m_2$  after collision.

  $v_{2f}=\left(\frac{m_2-m_1}{m_1+m_2}\right)v_{2i}+\left(\frac{2m_1}{m_1+m_2}\right)v_{1i}$                                                                       (IX)

Here,$v_{2f}$ is the velocity of the second skater after collision.

Equation (VII) can be used to calculate height maximum height attained by skaters if initial velocity is given.

Write the expression for the maximum height attained by first skater.

  $h_1=\frac{v_{1f}^2}{2g}$                                                                                                                (X)

Here, $h_1$ is the maximum height attained by skateboarder with mass $m_1$ .

Write the expression for the maximum height attained by second skater.

  $h_2=\frac{v_{2f}^2}{2g}$                                                                                                                 (XI)

Here, $h_2$ is the maximum height attained by skateboarder with mass $m_2$ .

Conclusion:

Substitute $75\text{kg}$ for $m_1$ , $65\text{kg}$ for $m_2$ , $8.86\text{m}/\text{s}$ for $v_{1i}$ and $-8.86\text{m}/\text{s}$ for $v_{2i}$ in equation (VIII) to get $v_{1f}$ .

  $\begin{array}{c}{v}_{1f}=\left(\frac{75\text{kg}-65\text{kg}}{75\text{kg}+65\text{kg}}\right)\left(8.86\text{m}/\text{s}\right)+\left(\frac{2\times 65\text{kg}}{75\text{kg}+65\text{kg}}\right)\left(-8.86\text{m}/\text{s}\right)\\ =-7.59\text{m}/\text{s}\end{array}$

Substitute $75\text{kg}$ for $m_1$ , $65\text{kg}$ for $m_2$ , $8.86\text{m}/\text{s}$ for $v_{1i}$ and $-8.86\text{m}/\text{s}$ for $v_{2i}$ in equation (IX) to get $v_{2f}$ .

  $\begin{array}{c}{v}_{2f}=\left(\frac{65\text{kg}-75\text{kg}}{75\text{kg}+65\text{kg}}\right)\left(8.86\text{m}/\text{s}\right)+\left(\frac{2\times 75\text{kg}}{75\text{kg}+65\text{kg}}\right)\left(-8.86\text{m}/\text{s}\right)\\ =10.1\text{m}/\text{s}\end{array}$

Substitute $-7.59\text{m}/\text{s}$ for $v_{1f}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (X) to get $h_1$ .

  $\begin{array}{c}{h}_1=\frac{{\left(-7.59\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ =2.94\text{m}\end{array}$

Substitute $10.1\text{m}/\text{s}$ for $v_{2f}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (XI) to get $h_2$ .

  $\begin{array}{c}{h}_2=\frac{{\left(10.1\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ =5.22\text{m}\end{array}$

Therefore, the height reached by skateboarder of mass $75.0\text{kg}$ after the collision is $2.94\text{m}$ and the height reached by skateboarder of mass $65.0\text{kg}$ after the collision is $5.22\text{m}$ .