Chapter 9, Problem 92A

a.

To determine

To Sketch:

Before and after situation of motion of woman and cart.

Explanation of Solution

Introduction:

Momentum of any body is equal to product of mass and velocity if any external force is not applied on the body then momentum of body is conserved.

Here, let the woman which is on the cart is moving in +x axis direction with a velocity of 5 m/s and after moving some distance woman jumps from the cart and velocity of woman is then 7 m/s in east ward direction.

This situation can could be sketched as follows:

  

b.

To determine

Velocity of cart when woman jumps off.

Answer to Problem 92A

Velocity of cart will be 5 m/s in westward direction.

Explanation of Solution

Given:

Mass of woman, $m_w=50\text{ kg}$

Mass of cart, $m_c=10\text{ kg}$

Common velocity of woman and cart, $v=5\text{ m/s}$

Velocity of woman, $v_w=7\text{ m/s}$

Formula used:

Momentum of any object is given by,

  $P=mv$

Here, P is momentum, m is mass and v is velocity

If no external force is applied on the object,then momentum is conserved.

  $P_i=P_f$

Calculation:

According to conservation of momentum, initial momentum will be equal to final momentum.

Therefore,

  $\begin{array}{l}{m}_w{v}_w+m_c{v}_c=\left(m_w+m_c\right)v\\ 50\times 7+10\times v_c=\left(50+10\right)\times 5\\ 10v_c+350=300\\ 10v_c=-50\\ v_c=-5\text{ m/s}\end{array}$

Here, negative sign indicates that velocity of cart will be in opposite direction of velocity of woman.

Conclusion:

Therefore, velocity of cart will be 5 m/s in westward direction.