To find the velocity of the ice skater after throwing the snowball, we can use the principle of conservation of momentum. According to this principle, the total momentum before the action must equal the total momentum after the action.
Let:
- \( m_s = 65.0 , \text{kg} \) (mass of the skater)
- \( m_b = 0.15 , \text{kg} \) (mass of the snowball)
- \( v_b = 32.0 , \text{m/s} \) (velocity of the snowball after being thrown)
- \( v_s \) = velocity of the skater after throwing the snowball (what we want to find)
Before the snowball is thrown, the total momentum is zero (since both the skater and the snowball are at rest). After throwing the snowball, the conservation of momentum gives us:
\[ (m_s \cdot 0) + (m_b \cdot 0) = m_s \cdot v_s + m_b \cdot v_b \]
This simplifies to:
\[ 0 = m_s \cdot v_s + m_b \cdot v_b \]
Rearranging the equation to solve for \( v_s \):
\[ m_s \cdot v_s = -m_b \cdot v_b \]
\[ v_s = -\frac{m_b \cdot v_b}{m_s} \]
Now plug in the values:
\[ v_s = -\frac{0.15 , \text{kg} \cdot 32.0 , \text{m/s}}{65.0 , \text{kg}} \]
Calculating \( v_s \):
\[ v_s = -\frac{4.8}{65.0} \]
\[ v_s \approx -0.0738 , \text{m/s} \]
Rounding to two decimal places, we find:
\[ v_s \approx -0.07 , \text{m/s} \]
So, the skater moves backward at approximately -0.07 m/s.
The answer is:
b: -0.07 m/s
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