Chapter 8, Problem 16TP

To determine

The combined velocity of the girl and platform after the jump, the combined momentum of the girl and platform both before and after the collision, momentum of the boy before and after the collision, to check whether the momentum is conserved or not and to state the one which is the example of an open system and the one which is the example of a closed system.

Answer to Problem 16TP

The combined velocity of the girl and platform after the jump is $1.25\text{m}/\text{s}$.

The combined momentum of the girl and platform both before and after the collision is $200\text{kg}\cdot \text{m}/\text{s}$.

Momentum of the boy before the collision is $300\text{kg}\cdot \text{m}/\text{s}$.

Momentum of the boy after the collision is $0\text{kg}\cdot \text{m}/\text{s}$.

The momentum is not conserved.

The case of the girl jumping on a platform is an example of closed system and the case of the boy colliding with a barrier is an example of open system.

Explanation of Solution

Given:

The mass of the girl is $m=40\text{kg}$.

The mass of the platform is $m=120\text{kg}$.

The initial speed of the girl is $v=5.0\text{m}/\text{s}$.

The mass of the boy is $m=50\text{kg}$.

The initial speed of the boy is $v=6.0\text{m}/\text{s}$.

Formula used:

The conservation of momentum for combined bodies is given by

  $m_1{v}_1+m_2{v}_2=m{v}^{\prime }$

Here, $m$ is the combined mass and $v^{\prime }$ is the combined velocity after the collision.

The combined mass is given by

  $m=m_1+m_2$

The combined momentum before collision is given by

  ${\left(\text{combined momentum}\right)}_{\text{before}}=m_1{v}_1+m_2{v}_2$

The combined momentum after collision is given by

  ${\left(\text{combined momentum}\right)}_{\text{after}}=mv$

The initial momentum is given by

  $p_{\text{i}}=m{v}_1$

The final momentum is given by

  $p_{\text{f}}=m{v}_2$

Calculation:

The combined mass is calculated as

  $\begin{array}{c}m=m_1+m_2\\ =40\text{kg}+120\text{kg}\\ =160\text{kg}\end{array}$

The combined velocity is calculated as

  $\begin{array}{c}v=\frac{m_1{v}_1+m_2{v}_2}{m}\\ =\frac{\left(40\text{kg}\right)\left(5.0\text{m}/\text{s}\right)+\left(120\text{kg}\right)0}{160\text{kg}}\\ =\frac{200\text{kg}\cdot \text{m}/\text{s}}{160\text{kg}}\\ =1.25\text{m}/\text{s}\end{array}$

The combined momentum before collision is calculated as

  $\begin{array}{c}{\left(\text{combined momentum}\right)}_{\text{before}}=m_1{v}_1+m_2{v}_2\\ =\left(40\text{kg}\right)\left(5.0\text{m}/\text{s}\right)+\left(120\text{kg}\right)0\\ =200\text{kg}\cdot \text{m}/\text{s}\end{array}$

The combined momentum after collision is calculated as

  $\begin{array}{c}{\left(\text{combined momentum}\right)}_{\text{after}}=mv\\ =\left(160\text{kg}\right)\left(1.25\text{m}/\text{s}\right)\\ =200\text{kg}\cdot \text{m}/\text{s}\end{array}$

The initial momentum of boy is calculated as

  $\begin{array}{c}{p}_{\text{i}}=m{v}_1\\ =\left(50\text{kg}\right)\left(6.0\text{m}/\text{s}\right)\\ =300\text{kg}\cdot \text{m}/\text{s}\end{array}$

The final momentum of the boy is calculated as

  $\begin{array}{c}{p}_{\text{f}}=m{v}_2\\ =\left(50\text{kg}\right)\left(0\text{m}/\text{s}\right)\\ =0\text{kg}\cdot \text{m}/\text{s}\end{array}$

Conclusion:

The combined velocity of the girl and platform after the jump is equal to $1.25\text{m}/\text{s}$.

The combined momentum of the girl and platform both before and after the collision is equal to $200\text{kg}\cdot \text{m}/\text{s}$.

Momentum of the boy before the collision is equal to $300\text{kg}\cdot \text{m}/\text{s}$.

Momentum of the boy after the collision is equal to $0\text{kg}\cdot \text{m}/\text{s}$.

The momentum of the system of boy and the barrier is not conserved, because the momentum of the boy before the collision is not equal to the momentum after the collision.

The case of the girl jumping on a platform is an example of closed system, because the girl and the platform start to swing like a pendulum after the girl jumps over the platform. Therefore, there is no net external force acting on them.

The case of the boy colliding with a barrier is an example of open system, because the wall is applying external force on the barrier and the boy.Due to this, the final momentum of the boy gets zero and the barrier does not gain any momentum.