A 35-kg girl is standing near and to the left of a 43-kg boy on the frictionless surface of a frozen pond. The boy throws a 0.75-kg ice ball to the girl with a horizontal speed of 6.2 m/s. What are the velocities of the boy and the girl immediately after the girl catches the ice ball?

Options

A. Girl - 0.81 m/s, left , Boy - 0.67 m/s , right
B. Girl - 0.71 m/s , left , Boy -0.41 m/s , left
C. Girl-0.18 m/s , right , Boy - 0.13 m/s , left
D. Girl-0.13 m/s , left , Boy - 0.11 m/s , right

Answer

To solve the problem, we will use the principle of conservation of momentum. The total momentum before the event (the boy throwing the ball) must equal the total momentum after the event (the girl catching the ball).

### Step 1: Calculate the initial momentum of the system
Before the boy throws the ball, both the boy and the girl are at rest, and the ball is moving towards the girl. The initial momentum of the system can be calculated as follows:

\[ \text{Initial momentum} = \text{momentum of the ball} + \text{momentum of the boy} + \text{momentum of the girl} \]

Since the boy and girl are initially at rest, their momenta are zero. Thus, we only consider the momentum of the ball:

\[ \text{Initial momentum} = m_{\text{ball}} \cdot v_{\text{ball}} = 0.75 \, \text{kg} \cdot 6.2 \, \text{m/s} = 4.65 \, \text{kg m/s} \]

### Step 2: Calculate the momentum after the girl catches the ball
After the girl catches the ball, the total mass of the girl and the ball combined is:

\[ m_{\text{girl}} + m_{\text{ball}} = 35 \, \text{kg} + 0.75 \, \text{kg} = 35.75 \, \text{kg} \]

Let \( v_g \) be the final velocity of the girl (and the ball) after catching it. The momentum after the catch can be expressed as:

\[ \text{Final momentum} = (m_{\text{girl}} + m_{\text{ball}}) \cdot v_g \]

### Step 3: Set initial momentum equal to final momentum
Using the conservation of momentum, we have:

\[ \text{Initial momentum} = \text{Final momentum} \]

Substituting the values we calculated:

\[ 4.65 \, \text{kg m/s} = 35.75 \, \text{kg} \cdot v_g \]

### Step 4: Solve for \( v_g \)
Now, we can solve for \( v_g \):

\[ v_g = \frac{4.65 \, \text{kg m/s}}{35.75 \, \text{kg}} \approx 0.13 \, \text{m/s} \]

### Step 5: Determine the direction of the velocities
Since the boy threw the ball to the right, the girl will move to the right as well after catching the ball.

### Step 6: Calculate the boy's velocity after throwing the ball
Now, we need to find the velocity of the boy after he throws the ball. By conservation of momentum, the momentum of the boy must also change when he throws the ball.

Let \( v_b \) be the velocity of the boy after throwing the ball:

\[ \text{Initial momentum of boy} = 0 \quad (\text{at rest before throwing}) \]

The momentum of the boy after throwing the ball is:

\[ \text{Final momentum of boy} = m_{\text{boy}} \cdot v_b \]

Setting the momentum of the boy equal to the momentum of the ball (in the opposite direction):

\[ 0 = 4.65 \, \text{kg m/s} - 43 \, \text{kg} \cdot v_b \]

Solving for \( v_b \):

\[ 43 \, \text{kg} \cdot v_b = 4.65 \, \text{kg m/s} \]
\[ v_b = \frac{4.65 \, \text{kg m/s}}{43 \, \text{kg}} \approx 0.11 \, \text{m/s} \]

### Final Velocities
- The girl's final velocity \( v_g \) is approximately \( 0.13 \, \text{m/s} \) to the right.
- The boy's final velocity \( v_b \) is approximately \( 0.11 \, \text{m/s} \) to the left.

### Summary
- Velocity of the girl after catching the ball: \( 0.13 \, \text{m/s} \) (to the right)
- Velocity of the boy after throwing the ball: \( 0.11 \, \text{m/s} \) (to the left)