Activity 12 A 35 kg giri is standing near and to the left of a 43 kg boy or the frictionless surface of a frozen pond. The boy throws a \( 0,75 \mathrm{~kg} \) ice ball to the girl with a horizontal speed of \( 6.2 \mathrm{~m} \cdot \mathrm{~s}^{-1} \). What are the velocities of the boy and the girl immediately after the girl catches the ice ball?

To solve this problem, we can use the principle of conservation of momentum. The total momentum before the girl catches the ice ball must equal the total momentum after she catches it. ### Known Conditions: - Mass of the girl, \( m_g = 35 \, \text{kg} \) - Mass of the boy, \( m_b = 43 \, \text{kg} \) - Mass of the ice ball, \( m_i = 0.75 \, \text{kg} \) - Initial speed of the ice ball, \( v_i = 6.2 \, \text{m/s} \) - Initial speed of the girl, \( v_{g_i} = 0 \, \text{m/s} \) (since she is standing still) - Initial speed of the boy, \( v_{b_i} = 0 \, \text{m/s} \) (since he is also standing still) ### Step 1: Calculate the initial momentum The initial momentum of the system (boy, girl, and ice ball) can be calculated as follows: \[ p_{\text{initial}} = m_b \cdot v_{b_i} + m_g \cdot v_{g_i} + m_i \cdot v_i \] Substituting the known values: \[ p_{\text{initial}} = 43 \cdot 0 + 35 \cdot 0 + 0.75 \cdot 6.2 \] ### Step 2: Calculate the final momentum After the girl catches the ice ball, the total mass of the girl and the ice ball is: \[ m_{g_f} = m_g + m_i \] The final momentum of the system can be expressed as: \[ p_{\text{final}} = (m_g + m_i) \cdot v_f + m_b \cdot v_{b_f} \] Where \( v_f \) is the final velocity of the girl (and the ice ball) and \( v_{b_f} \) is the final velocity of the boy (which remains 0 since he does not move). ### Step 3: Set initial momentum equal to final momentum Using the conservation of momentum: \[ p_{\text{initial}} = p_{\text{final}} \] Substituting the expressions we derived: \[ 0.75 \cdot 6.2 = (35 + 0.75) \cdot v_f + 43 \cdot 0 \] ### Step 4: Solve for \( v_f \) Now we can solve for \( v_f \): \[ 0.75 \cdot 6.2 = 35.75 \cdot v_f \] Calculating \( 0.75 \cdot 6.2 \): \[ 0.75 \cdot 6.2 = 4.65 \] Now we can solve for \( v_f \): \[ 4.65 = 35.75 \cdot v_f \] \[ v_f = \frac{4.65}{35.75} \] Now, let's calculate \( v_f \). Calculate the value by following steps: - step0: Calculate: \(\frac{4.65}{35.75}\) - step1: Convert the expressions: \(\frac{\frac{93}{20}}{\frac{143}{4}}\) - step2: Multiply by the reciprocal: \(\frac{93}{20}\times \frac{4}{143}\) - step3: Reduce the numbers: \(\frac{93}{5}\times \frac{1}{143}\) - step4: Multiply the fractions: \(\frac{93}{5\times 143}\) - step5: Multiply: \(\frac{93}{715}\) Calculating the final velocity of the girl after catching the ice ball gives us: \[ v_f \approx 0.133 \, \text{m/s} \] ### Summary of Results: - The final velocity of the girl (and the ice ball) after catching it is approximately \( 0.133 \, \text{m/s} \). - The boy's final velocity remains \( 0 \, \text{m/s} \) since he does not move. Thus, the velocities immediately after the girl catches the ice ball are: - **Girl's velocity**: \( 0.133 \, \text{m/s} \) - **Boy's velocity**: \( 0 \, \text{m/s} \)