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Chapter 7: Problem 78

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

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

Step by step solution

01

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 (\(v_{s_f}\)).
02

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 \(p_{total_{initial}} = m_sv_i + m_bv_i = (35.0 + 3.50)(5.00) = 192.5\,\text{kg}\cdot\text{m/s}\).
03

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 \(p_{final} = m_sv_{s_f} + m_bv_{b_f}\).
04

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.

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Most popular questions from this chapter

The nucleus of radioactive thorium- 228 , with a mass of about \(3.8 \cdot 10^{-25} \mathrm{~kg}\), is known to decay by emitting an alpha particle with a mass of about \(6.68 \cdot 10^{-27} \mathrm{~kg} .\) If the alpha particle is emitted with a speed of \(1.8 \cdot 10^{7} \mathrm{~m} / \mathrm{s},\) what is the recoil speed of the remaining nucleus (which is the nucleus of a radon atom)?

How fast would a \(5.00-\mathrm{g}\) fly have to be traveling to slow a \(1900 .-\mathrm{kg}\) car traveling at \(55.0 \mathrm{mph}\) by \(5.00 \mathrm{mph}\) if the fly hit the car in a totally inelastic head-on collision?

Rank the following objects from highest to lowest in terms of momentum and from highest to lowest in terms of energy. a) an asteroid with mass \(10^{6} \mathrm{~kg}\) and speed \(500 \mathrm{~m} / \mathrm{s}\) b) a high-speed train with a mass of \(180,000 \mathrm{~kg}\) and a speed of \(300 \mathrm{~km} / \mathrm{h}\) c) a 120 -kg linebacker with a speed of \(10 \mathrm{~m} / \mathrm{s}\) d) a 10 -kg cannonball with a speed of \(120 \mathrm{~m} / \mathrm{s}\) e) a proton with a mass of \(6 \cdot 10^{-27} \mathrm{~kg}\) and a speed of \(2 \cdot 10^{8} \mathrm{~m} / \mathrm{s}\).

Bats are extremely adept at catching insects in midair. If a 50.0-g bat flying in one direction at \(8.00 \mathrm{~m} / \mathrm{s}\) catches a \(5.00-\mathrm{g}\) insect flying in the opposite direction at \(6.00 \mathrm{~m} / \mathrm{s}\), what is the speed of the bat immediately after catching the insect?

A satellite with a mass of \(274 \mathrm{~kg}\) approaches a large planet at a speed \(v_{i, 1}=13.5 \mathrm{~km} / \mathrm{s}\). The planet is moving at a speed \(v_{i, 2}=10.5 \mathrm{~km} / \mathrm{s}\) in the opposite direction. The satellite partially orbits the planet and then moves away from the planet in a direction opposite to its original direction (see the figure). If this interaction is assumed to approximate an elastic collision in one dimension, what is the speed of the satellite after the collision? This so-called slingshot effect is often used to accelerate space probes for journeys to distance parts of the solar system (see Chapter 12).
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