Chapter 8, Problem 49P

(a)

To determine

The final velocity of the person and the cart.

Answer to Problem 49P

The final velocity of the person and the cart is ${\overrightarrow{v}}_f=1.33\text{m/s}\widehat{i}$.

Explanation of Solution

Write the expression for conservation of momentum for inelastic collision.

  $m_p{v}_i=\left(m_p+m_c\right)v_f$        (I)

Here, $m_p$ is the person, $v_i$ is the initial velocity of the person, $m_c$ is the mass of the cart, and $v_f$ is the final velocity of the person and the cart.

Conclusion:

Substitute $60\text{kg}$ for $m_p$, $4.00\text{m/s}$ for $v_i$, and $120\text{kg}$ for $m_c$ in the equation (I) to find $v_f$.

  $\begin{array}{c}\left(60\text{kg}\right)\left(4.00\text{m/s}\right)=\left(60\text{kg}+120\text{kg}\right)v_f\\ 240\text{kg}\cdot \text{m/s}=\left(180\text{kg}\right)v_f\\ v_f=1.33\text{m/s}\end{array}$

In vector notation the velocity of the cart and person is,

  ${\overrightarrow{v}}_f=1.33\text{m/s}\widehat{i}$

Therefore, the final velocity of the person and the cart is ${\overrightarrow{v}}_f=1.33\text{m/s}\widehat{i}$.

(b)

To determine

The friction force acting on the person.

Answer to Problem 49P

The friction force acting on the person is $-235\text{N}\widehat{i}$.

Explanation of Solution

Write the expression for normal force by using Newton’s second law in y direction

  $\begin{array}{c}{\displaystyle \sum F_y}=0\\ N-mg=0\end{array}$        (II)

Here, $N$ is the normal force exerted on the person, $m$ is the mass of the person, and $g$ is the acceleration due to gravity.

Write the expression for frictional force exerted on the person.

  $f_k={\mu }_{k}N$        (III)

Here, $f_k$ is the frictional force and ${\mu }_k$ is the coefficient of kinetic friction.

Conclusion:

Substitute $60\text{kg}$ for $m$ and $9.80{\text{m/s}}^2$ for $g$ in equation (II) to find $N$.

  $\begin{array}{c}N-\left(60\text{kg}\right)\left(9.80{\text{m/s}}^2\right)=0\\ N-588\text{N}=0\\ N=588\text{N}\end{array}$

Substitute $588\text{N}$ for $N$ and $0.400$ for ${\mu }_k$ in equation (III) to find $f_k$.

  $\begin{array}{c}{f}_k=\left(588\text{N}\right)\left(0.400\right)\\ =235\text{N}\end{array}$

In vector notation the frictional force is,

  ${\overrightarrow{f}}_k=-235\text{N}\widehat{i}$

Therefore, the friction force acting on the person is $-235\text{N}\widehat{i}$.

(c)

To determine

The time taken for the frictional force acting on the person.

Answer to Problem 49P

The time taken for the frictional force acting on the person is $0.680\text{s}$.

Explanation of Solution

Write the expression for person’s momentum equal to the impulse.

  $p_i+I=p_f$        (IV)

Here, $p_i$ is the initial momentum, $I$ is the impulse, and $p_f$ is the final momentum.

Write the expression for initial momentum.

  $p_i=m{v}_i$        (V)

Here, $m$ is the mass of the person and $v_i$ is the initial speed of the person.

Write the expression for final momentum.

  $p_f=m{v}_f$        (VI)

Here, $v_f$ is the final speed of the person and cart.

Write the expression for impulse.

  $I=Ft$        (VII)

Here, $F$ is the force exerted on the person and $t$ is the time taken for the frictional force acting on the person.

Conclusion:

Substitute the equations (V), (VI) and (VII) in equation (IV).

  $m{v}_i+Ft=m{v}_f$

Substitute $60\text{kg}$ for $m$, $4.00\text{m/s}$ for $v_i$, $-235\text{N}$ for $F$, and $1.33\text{m/s}$ for $v_f$ in above relation to find $t$.

  $\begin{array}{c}\left(60\text{kg}\right)\left(4.00\text{m/s}\right)-\left(235\text{N}\right)t=\left(60\text{kg}\right)\left(1.33\text{m/s}\right)\\ 240\text{kg}\cdot \text{m/s}-\left(235\text{N}\right)t=79.8\text{kg}\cdot \text{m/s}\\ -\left(235\text{N}\right)t=79.8\text{kg}\cdot \text{m/s}-240\text{kg}\cdot \text{m/s}\\ t=0.680\text{s}\end{array}$

Therefore, the time taken for the frictional force acting on the person is $0.680\text{s}$.

(d)

To determine

The change in momentum of the person and the cart.

Answer to Problem 49P

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

Explanation of Solution

Write the expression for change in momentum of the person.

  $p_p=m\left({\overrightarrow{v}}_f-{\overrightarrow{v}}_i\right)$        (VIII)

Write the expression for change in momentum of the cart.

  $p_c=m\left({\overrightarrow{v}}_f-{\overrightarrow{v}}_i\right)$        (IX)

Conclusion:

Substitute $60\text{kg}$ for $m$, $4.00\text{m/s}$ for $v_i$, and $1.33\text{m/s}$ for $v_f$ in equation (VIII) to find $p_p$.

  $\begin{array}{c}{p}_p=60\text{kg}\left(1.33\text{m/s}-4.00\text{m/s}\right)\\ =-160\text{N}\cdot \text{s}\widehat{i}\end{array}$

Substitute $120\text{kg}$ for $m$, $0\text{m/s}$ for $v_i$, and $1.33\text{m/s}$ for $v_f$ in equation (IX) to find $p_c$.

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

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

(e)

To determine

The displacement of the person relative to the ground.

Answer to Problem 49P

The displacement of the person relative to the ground is $1.81\text{m}$.

Explanation of Solution

Write the expression for displacement of the person relative to the ground.

  $x_f-x_i=\frac{1}{2}\left(v_i+v_f\right)t$        (X)

Here, $x_i$ is the initial distance and $x_f$ is the final distance.

Conclusion:

Substitute $4.00\text{m/s}$ for $v_i$, $0.680\text{s}$ for $t$, and $1.33\text{m/s}$ for $v_f$ in equation (X) to find $x_f-x_i$.

  $\begin{array}{c}{x}_f-x_i=\frac{1}{2}\left(4.00\text{m/s}+1.33\text{m/s}\right)\left(0.680\text{s}\right)\\ =1.81\text{m}\end{array}$

Therefore, the displacement of the person relative to the ground is $1.81\text{m}$.

(f)

To determine

The displacement of the cart relative to the ground.

Answer to Problem 49P

The displacement of the cart relative to the ground is $0.454\text{m}$.

Explanation of Solution

Write the expression for displacement of the cart relative to the ground.

  $x_f-x_i=\frac{1}{2}\left(v_i+v_f\right)t$        (XI)

Conclusion:

Substitute $0\text{m/s}$ for $v_i$, $0.680\text{s}$ for $t$, and $1.33\text{m/s}$ for $v_f$ in equation (XI) to find $x_f-x_i$.

  $\begin{array}{c}{x}_f-x_i=\frac{1}{2}\left(0\text{m/s}+1.33\text{m/s}\right)\left(0.680\text{s}\right)\\ =0.454\text{m}\end{array}$

Therefore, the displacement of the cart relative to the ground is $0.454\text{m}$.

(g)

To determine

The change in kinetic energy of the person.

Answer to Problem 49P

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

Explanation of Solution

Write the expression for change in kinetic energy of the person.

  $\Delta K_p=\frac{1}{2}m\left(v_f^2-v_i^2\right)$        (XII)

Conclusion:

Substitute $4.00\text{m/s}$ for $v_i$, $60\text{kg}$ for $m$, and $1.33\text{m/s}$ for $v_f$ in equation (XII) to find $\Delta K_p$.

  $\begin{array}{c}\Delta K_p=\frac{1}{2}60\text{kg}\left({\left(1.33\text{m/s}\right)}^2-{\left(4.00\text{m/s}\right)}^2\right)\\ =-427\text{J}\end{array}$

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

(h)

To determine

The change in kinetic energy of the cart.

Answer to Problem 49P

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

Explanation of Solution

Write the expression for change in kinetic energy of the cart.

  $\Delta K_c=\frac{1}{2}m\left(v_f^2-v_i^2\right)$        (XIII)

Conclusion:

Substitute $0\text{m/s}$ for $v_i$, $120\text{kg}$ for $m$, and $1.33\text{m/s}$ for $v_f$ in equation (XIII) to find $\Delta K_c$.

  $\begin{array}{c}\Delta K_c=\frac{1}{2}120\text{kg}\left({\left(1.33\text{m/s}\right)}^2-{\left(0\text{m/s}\right)}^2\right)\\ =107\text{J}\end{array}$

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

(i)

To determine

Why the answers part (g) and (h) are differ.

Answer to Problem 49P

Because. the distance moved by the cart is different from the distance moved by the point of application of friction force to the cart.

Explanation of Solution

The force exerted by the person on the cart must be equal in magnitude and opposite in direction to the force exerted by the cart on the person. The changes in momentum of the two objects must be equal in magnitude and must be added to zero.

The change in kinetic energy is different in magnitude and does not add to zero.

Conclusion:

The following situation is represents in two ways,

The distance moved by the cart is different from the distance moved by the point of application of friction force to the cart.

The total change in mechanical energy for both objects add together becomes zero, it is perfectly in elastic collision.