Question

A skateboarder of mass \(35.0 \mathrm{~kg}\) is riding her \(3.50-\mathrm{kg}\) skateboard at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). She jumps backward off her skateboard, sending the skateboard forward at a speed of \(8.50 \mathrm{~m} / \mathrm{s}\). At what speed is the skateboarder moving when her feet hit the ground?

Short answer

Answer: The final velocity of the skateboarder is approximately 4.64 m/s.

Step-by-step solution

Identify the initial data

The initial data provided in the problem are the mass of the skateboarder (\(m_s = 35.0\,\text{kg}\)), the mass of the skateboard (\(m_b = 3.50\,\text{kg}\)), the initial velocity of the skateboarder and skateboard (\(v_i = 5.00\,\text{m/s}\)), and the final velocity of the skateboard (\(v_{b_f} = 8.50\,\text{m/s}\)). Our goal is to find the final velocity of the skateboarder (\(v_{s_f}\)).

Calculate the initial total momentum

Calculate the initial total momentum (before the skateboarder jumps off) using the equation for momentum, \(p = mv\). The total initial momentum is the combined momentum of the skateboarder and the skateboard, so \(p_{total_{initial}} = m_sv_i + m_bv_i = (35.0 + 3.50)(5.00) = 192.5\,\text{kg}\cdot\text{m/s}\).

Applying the conservation of momentum

According to the conservation of momentum principle, the total momentum before the jump equals the total momentum after the jump. So, \(p_{total_{initial}} = p_{total_{final}}\). Now, we need to find the final total momentum. After the jump, the momentum of the skateboard and the skateboarder is \(p_{final} = m_sv_{s_f} + m_bv_{b_f}\).

Calculate the final velocity of the skateboarder

Now we can set up an equation to solve for the final velocity of the skateboarder: \(192.5 = 35.0v_{s_f} + 3.50(8.50)\). Simplify the equation and solve for \(v_{s_f}\): \[192.5 = 35.0v_{s_f} + 29.75\] \[v_{s_f} = \frac{192.5-29.75}{35.0} \approx 4.64\, \text{m/s}\] The skateboarder is moving at a speed of \(4.64\,\text{m/s}\) when her feet hit the ground.