Chapter 9, Problem 9.69AP

Review. A 60.0-kg person running at an initial speed of 4.00 m/s jumps onto a 120-kg cart initially at rest (Fig. P9.37). The person slides on the cart’s top surface and finally comes to rest relative to the cart. The coefficient of kinetic friction between the person and the cart is 0.400. Friction between the cart and ground can be ignored. (a) Find the final velocity of the person and cart relative to the ground. (b) Find the friction force acting on the person while he is sliding across the top surface of the cart. (c) How long does the friction force act on the person? (d) Find the change in momentum of the person and the change in momentum of the cart. (c) Determine the displacement of the person relative to the ground while he is sliding on the cart. (f) Determine the displacement of the cart relative to the ground while the person is sliding. (g) Find the change in kinetic energy of the person. (h) Find the change in kinetic energy of the cart. (i) Explain why the answers to (g) and (h) differ. (What kind of collision is this one, and what accounts for the loss of mechanical energy)

Figure P9.37

To determine

The final velocity of the person and cart relative to the ground.

Answer to Problem 9.69AP

The final velocity of the person and cart relative to the ground is $1.33\widehat{\text{i}}\text{m}/\text{s}$.

Explanation of Solution

The mass of the person is $60.0\text{kg}$, and the initial speed is $4.00\text{m}/\text{s}$. The mass of the cart is $120\text{kg}$ and the cart is at rest. The coefficient of kinetic friction between the cart and the person is $0.400$.

Write the expression of conservation of momentum.

    $m_{\text{p}}{v}_{\text{i}}=\left(m_{\text{p}}+m_{\text{c}}\right)v_{\text{f}}$

Here, $m_{\text{p}}$ is the mass of the person, $m_{\text{c}}$ is the mass of the cart, $v_{\text{i}}$ is the initial velocity of the person and $v_{\text{f}}$ is the final velocity of the person and cart relative to the ground.

Substitute $60.0\text{kg}$ for $m_{\text{p}}$, $4.00\text{m}/\text{s}$ for $v_{\text{i}}$ and $120\text{kg}$ for $m_{\text{c}}$ in above equation to find $v_{\text{f}}$.

  $\begin{array}{c}60.0\text{kg}\times 4.00\text{m}/\text{s}=\left(60.0\text{kg}+120\text{kg}\right)v_{\text{f}}\\ v_{\text{f}}=1.33\text{m}/\text{s}\\ =1.33\widehat{\text{i}}\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the final velocity of the person and cart relative to the ground is $1.33\widehat{\text{i}}\text{m}/\text{s}$.

To determine

The frictional force acting on the person while he is sliding.

Answer to Problem 9.69AP

The frictional force acting on the person while he is sliding is $-\text{235}\widehat{\text{i}}\text{N}$.

Explanation of Solution

Write the expression to calculate the frictional force.

    $f_{\text{k}}=\mu m_{\text{p}}g$

Here, $\mu$ is the coefficient of kinetic friction between the cart and the person and $g$ is the acceleration due to gravity.

Substitute $0.400$ for $\mu$, $60.0\text{kg}$ for $m_{\text{p}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in above equation to find $f_{\text{k}}$.

    $\begin{array}{c}{f}_{\text{k}}=0.400\times 60.0\text{kg}\times 9.81\text{m}/{\text{s}}^{\text{2}}\\ =235\text{N}\\ =-\text{235}\widehat{\text{i}}\text{N}\end{array}$

The negative sign indicates that the frictional force is acting toward negative x axis.

Conclusion:

Therefore, the frictional force acting on the person while he is sliding is $-\text{235}\widehat{\text{i}}\text{N}$.

To determine

The time duration in which the frictional force is acting on the person.

Answer to Problem 9.69AP

The time duration in which the frictional force is acting on the person is $0.680\text{s}$.

Explanation of Solution

Write the expression of Impulse-momentum equation.

    $f_{\text{k}}\Delta t=m_{\text{p}}\left(v_{\text{f}}-v_{\text{i}}\right)$

Here, $\Delta t$ is the time duration in which the frictional force is acting on the person.

Substitute $-235\text{N}$ for $f_{\text{k}}$, $1.33\text{m}/\text{s}$ for $v_{\text{f}}$, $4.00\text{m}/\text{s}$ for $v_{\text{i}}$ and $60.0\text{kg}$ for $m_{\text{p}}$ in above equation to find $\Delta t$.

    $\begin{array}{c}-235\text{N}\times \Delta t=60.0\text{kg}\times \left(1.33\text{m}/\text{s}-4.00\text{m}/\text{s}\right)\\ \Delta t=0.680\text{s}\end{array}$

Conclusion:

Therefore, the time duration in which the frictional force is acting on the person is $0.680\text{s}$.

To determine

The change in momentum of the person and cart.

Answer to Problem 9.69AP

The change in momentum of the person and cart is $-160\widehat{\text{i}}\text{N}\cdot \text{s}$ and $160\widehat{\text{i}}\text{N}\cdot \text{s}$ respectively.

Explanation of Solution

Write the expression to calculate the change in momentum of the person.

    $\Delta p_{\text{p}}=m_{\text{p}}\left(v_{\text{f}}-v_{\text{i}}\right)$

Substitute $1.33\widehat{\text{i}}\text{m}/\text{s}$ for $v_{\text{f}}$, $4.00\widehat{\text{i}}\text{m}/\text{s}$ for $v_{\text{i}}$ and $60.0\text{kg}$ for $m_{\text{p}}$ in above equation to find $\Delta p_{\text{p}}$.

    $\begin{array}{c}\Delta p_{\text{p}}=60.0\text{kg}\times \left(1.33-4.00\right)\widehat{\text{i}}\text{m}/\text{s}\\ \approx -160\widehat{\text{i}}\text{N}\cdot \text{s}\end{array}$

Write the expression to calculate the change in momentum of the cart.

    $\Delta p_{\text{c}}=m_{\text{c}}\left(v_{\text{f}}-v_{\text{i}}\right)$

Substitute $1.33\widehat{\text{i}}\text{m}/\text{s}$ for $v_{\text{f}}$, $0$ for $v_{\text{i}}$ and $120\text{kg}$ for $m_{\text{c}}$ in above equation to find $\Delta p_{\text{c}}$.

    $\begin{array}{c}\Delta p_{\text{p}}=120\text{kg}\times \left(1.33-0\right)\widehat{\text{i}}\text{m}/\text{s}\\ \approx 160\widehat{\text{i}}\text{N}\cdot \text{s}\end{array}$

Conclusion:

Therefore, the change in momentum of the person and cart is $-160\widehat{\text{i}}\text{N}\cdot \text{s}$.and $160\widehat{\text{i}}\text{N}\cdot \text{s}$ respectively.

To determine

The displacement of the person relative to the ground during sliding on the cart.

Answer to Problem 9.69AP

The displacement of the person relative to the ground during sliding on the cart is $\text{1}\text{.81}\text{m}$.

Explanation of Solution

Write the expression to calculate the displacement of the person.

    $\Delta d_{\text{p}}=\frac{1}{2}\left(v_{\text{i}}+v_{\text{f}}\right)\Delta t$

Substitute $1.33\widehat{\text{i}}\text{m}/\text{s}$ for $v_{\text{f}}$, $4.00\widehat{\text{i}}\text{m}/\text{s}$ for $v_{\text{i}}$ and $0.680\text{s}$ for $\Delta t$ in above equation to find $\Delta d_{\text{p}}$.

    $\begin{array}{c}\Delta d_{\text{p}}=\frac{1}{2}\left(1.33\widehat{\text{i}}\text{m}/\text{s}+4.00\widehat{\text{i}}\text{m}/\text{s}\right)\times 0.680\text{s}\\ \text{=1}\text{.81}\text{m}\end{array}$

Conclusion:

Therefore, the displacement of the person relative to the ground during sliding on the cart is $\text{1}\text{.81}\text{m}$.

To determine

The displacement of the cart relative to the ground during the person sliding on the cart.

Answer to Problem 9.69AP

The displacement of the cart relative to the ground during the person sliding on the cart is $\text{0}\text{.454}\text{m}$.

Explanation of Solution

Write the expression to calculate the displacement of the cart.

    $\Delta d_{\text{c}}=\frac{1}{2}\left(v_{\text{i}}+v_{\text{f}}\right)\Delta t$

Substitute $1.33\widehat{\text{i}}\text{m}/\text{s}$ for $v_{\text{f}}$, $0$ for $v_{\text{i}}$ and $0.680\text{s}$ for $\Delta t$ in above equation to find $\Delta d_{\text{c}}$.

    $\begin{array}{c}\Delta d_{\text{c}}=\frac{1}{2}\left(1.33\widehat{\text{i}}\text{m}/\text{s}+0\right)\times 0.680\text{s}\\ \text{=0}\text{.454}\text{m}\end{array}$

Conclusion:

Therefore, the displacement of the cart relative to the ground during the person sliding on the cart is $\text{0}\text{.454}\text{m}$.

To determine

The change in kinetic energy of the person.

Answer to Problem 9.69AP

The change in kinetic energy of the person is $-427\text{J}$.

Explanation of Solution

Write the expression to calculate the change in kinetic energy of the person.

    $\Delta K{E}_{\text{P}}=\frac{1}{2}{m}_{\text{p}}\left(v_{\text{f}}^{\text{2}}-v_{\text{i}}^{\text{2}}\right)$

Substitute $1.33\text{m}/\text{s}$ for $v_{\text{f}}$, $4.00\text{m}/\text{s}$ for $v_{\text{i}}$ and $60.0\text{kg}$ for $m_{\text{p}}$ in above equation to find $\Delta K{E}_{\text{P}}$.

`    $\begin{array}{c}\Delta K{E}_{\text{P}}=\frac{1}{2}\times 60.0\text{kg}\times \left[{\left(1.33\text{m}/\text{s}\right)}^2-{\left(4.00\text{m}/\text{s}\right)}^2\right]\\ =-427\text{J}\end{array}$

Conclusion:

Therefore, the change in kinetic energy of the person is $-427\text{J}$.

To determine

The change in kinetic energy of the cart.

Answer to Problem 9.69AP

The change in kinetic energy of the cart is $107\text{J}$.

Explanation of Solution

Write the expression to calculate the change in kinetic energy of the cart.

    $\Delta K{E}_{\text{c}}=\frac{1}{2}{m}_{\text{c}}\left(v_{\text{f}}^{\text{2}}-v_{\text{i}}^{\text{2}}\right)$

Substitute $1.33\text{m}/\text{s}$ for $v_{\text{f}}$, $0$ for $v_{\text{i}}$ and $120\text{kg}$ for $m_{\text{c}}$ in above equation to find $\Delta K{E}_{\text{c}}$.

`    $\begin{array}{c}\Delta K{E}_{\text{c}}=\frac{1}{2}\times 120\text{kg}\times \left[{\left(1.33\text{m}/\text{s}\right)}^2-0\right]\\ \approx 107\text{J}\end{array}$

Conclusion:

Therefore, the change in kinetic energy of the cart is $107\text{J}$.

To determine

The reason due to which the answer in part (g) and (h) are different.

Answer to Problem 9.69AP

The collision between the person and the cart is perfectly inelastic collision and the loss of energy is due to frictional force.

Explanation of Solution

The force acting on the person must be equal in magnitude and opposite in direction to the force exerted by the cart on the person.

According to the conservation of linear momentum, the change in momentum of the person and the cart must be equal in magnitude and must add to zero. The change in kinetic energy of the person and cart must be equal for elastic collision but in this case change in kinetic energy of the person and cart is not equal, which shows that this is inelastic collision.

The reason of change in kinetic energy of both object of not being same is, the displacement of person and the cart is not same due to frictional force acting on the person.

Due to the frictional force the loss of energy in form of heat is the internal energy.

Write the expression to calculate the internal energy.

    $E=K{E}_{\text{p}}-K{E}_{\text{c}}$

Substitute $-427\text{J}$ for $K{E}_{\text{p}}$ and $107\text{J}$ for $K{E}_{\text{c}}$ in above equation.

    $\begin{array}{c}E=427\text{J}-107\text{J}\\ \text{=320}\text{J}\end{array}$

Thus, the collision between the person and the cart is perfectly inelastic collision due to the loss of energy.

Conclusion:

Therefore, the collision between the person and the cart is perfectly inelastic collision and the loss of energy is due to frictional force.