Sheila weighs 60 kg and is riding a bike. Her momentum on the bike is 340 kg • m/s. The bike hits a rock, which stops it completely and throws Sheila forward onto the pavement.

If there is no net force on the system, what is Sheila’s velocity immediately after she is thrown from the bike?

1.8 m/s
2.0 m/s
5.0 m/s
5.7 m/s

1 answer

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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