Question

two ice skaters standing still, one with mass 50 kg and the other with mass 70 kg, push off from each other. if the 50 kg skater moves at 3 m/s, what is the velocity of the 70 kg skater? $m_1v_{i1} + m_2v_{i2} = m_1v_{f1} + m_2v_{f2}$
options: 2.14 m/s, -3.0 m/s, 3.0 m/s, -2.14 m/s

Explanation:

Step1: Set up conservation of momentum

Since both skaters start at rest, initial total momentum is 0. The formula is:
$$m_1v_{i1} + m_2v_{i2} = m_1v_{f1} + m_2v_{f2}$$
Substitute initial velocities ($v_{i1}=0, v_{i2}=0$):
$$0 = m_1v_{f1} + m_2v_{f2}$$

Step2: Rearrange for unknown velocity

Isolate $v_{f2}$ (velocity of 70 kg skater):
$$v_{f2} = -\frac{m_1v_{f1}}{m_2}$$

Step3: Substitute given values

$m_1=50\ \text{kg}$, $v_{f1}=3\ \text{m/s}$, $m_2=70\ \text{kg}$:
$$v_{f2} = -\frac{50 \times 3}{70} = -\frac{150}{70} \approx -2.14\ \text{m/s}$$
The negative sign indicates opposite direction to the 50 kg skater.

Answer:

-2.14 m/s