Two ice skaters want to prove conservation of momentum skater a has a mass of 72 KG and skater B has a mass of 48 KG they are originally at rest they both hold hands and face each other. Then they push each other and move in opposite directions skater moves with a velocity of 5MS what must the velocity of skater a B if the opposite direction after the push in order to prove conservation of momentum

Answer 1

To prove conservation of momentum, we need to ensure that the total momentum before and after the skaters push off from each other is equal.

Initial momentum before the push:

• Since both skaters are at rest, their initial momentum is: \[ p_{\text{initial}} = 0 \]

After the push, let's denote:

• The mass of Skater A (\(m_A\)) = 72 kg
• The mass of Skater B (\(m_B\)) = 48 kg
• The velocity of Skater A after the push (\(v_A\)) = 5 m/s (to the right)
• The velocity of Skater B after the push (\(v_B\)) (to the left, so it will be negative in our calculations).

Using conservation of momentum, the total momentum after the push should also equal zero: \[ p_{\text{final}} = m_A \cdot v_A + m_B \cdot v_B \] Substituting in the values we have: \[ 0 = (72 , \text{kg} \cdot 5 , \text{m/s}) + (48 , \text{kg} \cdot v_B) \] Solving for \(v_B\): \[ 0 = 360 + 48 v_B \] \[ 48 v_B = -360 \] \[ v_B = -\frac{360}{48} \] \[ v_B = -7.5 , \text{m/s} \]

So, Skater B must move at a velocity of 7.5 m/s in the opposite direction after the push to conserve momentum.