To find Sheila's velocity immediately after being thrown from the bike, we can use the concept of momentum conservation.
Initially, the total momentum of the system (Sheila plus the bike) is given as 340 kg·m/s. When the bike hits the rock and comes to a complete stop, its momentum becomes 0. The momentum is conserved, meaning the momentum before the collision must equal the momentum after it.
Let’s denote:
- \( m_S \) = mass of Sheila = 60 kg
- \( v_S \) = velocity of Sheila after being thrown
- The initial momentum \( p_i \) is 340 kg·m/s.
- The final momentum \( p_f \) must equal \( p_i \).
Using the equation for momentum: \[ p_i = p_f = m_S \cdot v_S \] Since the bike stops and does not contribute to the final momentum, we have: \[ 340 , \text{kg·m/s} = 60 , \text{kg} \cdot v_S \]
Now we can solve for \( v_S \): \[ v_S = \frac{340 , \text{kg·m/s}}{60 , \text{kg}} = \frac{340}{60} \approx 5.67 , \text{m/s} \]
Rounding this to one decimal place, we get: \[ v_S \approx 5.7 , \text{m/s} \]
Thus, Sheila’s velocity immediately after being thrown from the bike is approximately 5.7 m/s.
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