Question

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.