Question

On an ice skating rink, a girl of mass 50 kg stands stationary, face to face with a boy of mass 80 kg. The children push off of one another, and the boy moves away with a velocity of +3 m/s.

What is the final velocity of the girl?

–1.9 m/s
+1.9 m/s
–4.8 m/s
+4.8 m/s

Answer 1

To solve for the final velocity of the girl after the boy pushes off from her, we can use the principle of conservation of momentum. According to this principle, the total momentum before the push must equal the total momentum after the push.

Let:
- Mass of the girl \( m_g = 50 \) kg
- Mass of the boy \( m_b = 80 \) kg
- Final velocity of the girl \( v_g \)
- Final velocity of the boy \( v_b = +3 \) m/s (given)

Before the push, both the girl and the boy are stationary, so the total initial momentum \( p_{initial} \) is 0:

\[
p_{initial} = m_g \cdot 0 + m_b \cdot 0 = 0
\]

After they push off from each other, their momenta will be:

\[
p_{final} = m_g \cdot v_g + m_b \cdot v_b
\]

Setting the initial momentum equal to the final momentum, we have:

\[
0 = (50 \, \text{kg}) \cdot v_g + (80 \, \text{kg}) \cdot (+3\, \text{m/s})
\]

Now we can solve for \( v_g \):

\[
0 = 50 v_g + 240
\]

\[
50 v_g = -240
\]

\[
v_g = \frac{-240}{50}
\]

\[
v_g = -4.8 \, \text{m/s}
\]

So, the final velocity of the girl is \(-4.8 \, \text{m/s}\).

The correct answer is:

–4.8 m/s