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Chapter 3: Problem 4

Two hobos, each of mass \(m_{\mathrm{h}}\), are standing at one end of a stationary railroad flatcar with frictionless wheels and mass \(m_{\mathrm{fc}} .\) Either hobo can run to the other end of the flatcar and jump off with the same speed \(u\) (relative to the car). (a) Use conservation of momentum to find the speed of the recoiling car if the two men run and jump simultaneously. (b) What is it if the second man starts running only after the first has already jumped? Which procedure gives the greater speed to the car? [Hint: The speed \(u\) is the speed of either hobo, relative to the car just after he has jumped; it has the same value for either man and is the same in parts (a) and (b).]

Short Answer

Expert verified
Both procedures result in the same car speed.

Step by step solution

01

Understand the System and Concepts

The problem can be approached using the conservation of momentum. Initially, the system is at rest, so the total momentum is zero. As the hobos jump, they set the flatcar into motion, and we need to calculate the resulting speed of the car under two scenarios: both hobos jump simultaneously, and the hobos jump one after the other.
02

Apply Momentum Conservation for Simultaneous Jump

For the first scenario, when both hobos jump simultaneously, the momentum of the system initially is zero. When they jump, each hobo moves with speed \(u\) relative to the car. If the car moves with speed \(v_{\text{fc}}\) in the opposite direction to conserve momentum, we have:\[2 m_{\text{h}} u = (2 m_{\text{h}} + m_{\text{fc}}) v_{\text{fc}}\]Solving for \(v_{\text{fc}}\):\[v_{\text{fc}} = \frac{2 m_{\text{h}} u}{2 m_{\text{h}} + m_{\text{fc}}}\]
03

Apply Momentum Conservation for Successive Jumps

In the second scenario, one hobo jumps, and then the other. When the first hobo jumps, the momentum equation is:\[m_{\text{h}} u = (m_{\text{h}} + m_{\text{fc}}) v_1\]Solving for \(v_1\), the velocity of the car after the first jump:\[v_1 = \frac{m_{\text{h}} u}{m_{\text{h}} + m_{\text{fc}}}\]Now, as the second hobo jumps, the flatcar already has speed \(v_1\). The second jump's momentum equation:\[m_{\text{h}} (u + v_1) = m_{\text{fc}} v_2 + m_{\text{h}} v_2\]Solve for \(v_2\), with \(v_2\) as the final velocity of the car:\[v_2 = \frac{m_{\text{h}} (u + \frac{m_{\text{h}} u}{m_{\text{h}} + m_{\text{fc}}})}{m_{\text{fc}}}\]Simplifying, \[v_2 = \frac{2 m_{\text{h}} u}{(m_{\text{h}} + m_{\text{fc}})}\]
04

Compare Both Procedures

By calculating the momentum equations in both scenarios, we can conclude that the speed of the car, \(v_{\text{fc}}\) in both situations is the same.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Simultaneous Jump
When two hobos jump off the flatcar at the same time, the system experiences a very specific dynamic due to the principle of conservation of momentum. Initially, everything is at rest, so the total momentum is zero. By the time the hobos jump, they propel themselves in one direction, while the flatcar gets an equal and opposite momentum. This is in accordance with Newton's Third Law and the conservation of momentum.

The formula that describes this balance of momentum for a simultaneous jump is:\[2 m_\text{h} u = (2 m_\text{h} + m_\text{fc}) v_\text{fc}\]Where:
  • \( m_\text{h} \) is the mass of each hobo.
  • \( u \) is the speed of each hobo relative to the car.
  • \( m_\text{fc} \) is the mass of the flatcar.
  • \( v_\text{fc} \) is the final speed of the flatcar.
By solving for \( v_\text{fc} \), we find the flatcar's recoil speed:
\[v_\text{fc} = \frac{2 m_\text{h} u}{2 m_\text{h} + m_\text{fc}}\]This shows how both hobos jumping simultaneously results in a shared push that sets the flatcar in motion, illustrating how combined kinetic actions translate into movement according to the principles of momentum.
Successive Jumps
When the hobos decide to jump off one after the other, the sequence of actions involves a series of momentum transfer that builds over time. Initially, when the first hobo jumps, the momentum caused by his jump pushes the flatcar backward, granting it a certain speed \(v_1\).

This initial momentum transfer is represented by the equation:\[m_\text{h} u = (m_\text{h} + m_\text{fc}) v_1\]By solving for \(v_1\), we determine the car鈥檚 speed following the first jump:
\[v_1 = \frac{m_\text{h} u}{m_\text{h} + m_\text{fc}}\]With the second hobo jumping, the scenario changes slightly. Now, the flatcar already has some velocity \(v_1\), so the second jump builds on this initial movement:\[m_\text{h} (u + v_1) = m_\text{fc} v_2 + m_\text{h} v_2\]\[v_2 = \frac{2 m_\text{h} u}{m_\text{h} + m_\text{fc}}\]This final velocity \(v_2\) demonstrates how step-by-step actions lead to reaching the same conclusion as simultaneous actions, although through a different series of events.
Momentum Equations
The conservation of momentum is a fundamental principle that helps us understand how motion is transferred in a system. At its core, the idea is that the total momentum of an isolated system remains constant if it is not subjected to external forces. Applied in the context of the hobos and the flatcar, each jump alters the momentum distribution within the system, but the total momentum remains unchanged.

In both simultaneous and successive jumps, the equations are built around:
  • Total initial momentum = Total final momentum
  • System momentum starts at zero when at rest
  • As actions occur (jumps), internal forces shift momentum
By applying momentum conservation to both jumping scenarios, we derive equations that account for the hobos' effect on the flatcar's momentum. This highlights how:
  • Simultaneous jumps consolidate impulses into one reaction
  • Successive jumps manage each impulse uniquely
Despite the paths being different, these momentum equations stress the robustness of conservation laws in physics, leading to the same observable outcomes for the flatcar鈥檚 final speed.

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Most popular questions from this chapter

A rocket (initial mass \(m_{\mathrm{o}}\) ) needs to use its engines to hover stationary, just above the ground. (a) If it can afford to burn no more than a mass \(\lambda m_{\mathrm{o}}\) of its fuel, for how long can it hover? [Hint: Write down the condition that the thrust just balance the force of gravity. You can integrate the resulting equation by separating the variables \(t\) and \(m .\) Take \(v_{\text {ex }}\) to be constant.] (b) If \(v_{\mathrm{ex}} \approx 3000 \mathrm{m} / \mathrm{s}\) and \(\lambda \approx 10 \%,\) for how long could the rocket hover just above the earth's surface?

Consider a gun of mass \(M\) (when unloaded) that fires a shell of mass \(m\) with muzzle speed \(v\). (That is, the shell's speed relative to the gun is \(v\).) Assuming that the gun is completely free to recoil (no external forces on gun or shell), use conservation of momentum to show that the shell's speed relative to the ground is \(v /(1+m / M).\)

A shell traveling with speed \(v_{\mathrm{o}}\) exactly horizontally and due north explodes into two equal-mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed \(v_{\mathrm{o}} .\) What is the velocity of the other fragment?

Show that the moment of inertia of a uniform solid sphere rotating about a diameter is \(\frac{2}{5} M R^{2}\). The sum (3.31) must be replaced by an integral, which is easiest in spherical polar coordinates, with the axis of rotation taken to be the \(z\) axis. The element of volume is \(d V=r^{2} d r \sin \theta d \theta d \phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)

Consider a uniform solid disk of mass \(M\) and radius \(R\), rolling without slipping down an incline which is at angle \(\gamma\) to the horizontal. The instantaneous point of contact between the disk and the incline is called \(P\). (a) Draw a free-body diagram, showing all forces on the disk. (b) Find the linear acceleration \(\dot{v}\) of the disk by applying the result \(\dot{\mathbf{L}}=\Gamma^{\text {ext }}\) for rotation about \(P .\) (Remember that \(L=I \omega\) and the moment of inertia for rotation about a point on the circumference is \(\frac{3}{2} M R^{2} .\) The condition that the disk not slip is that \(v=R \omega \text { and hence } \dot{v}=R \dot{\omega} .)\) (c) Derive the same result by applying \(\dot{\mathbf{L}}=\mathbf{\Gamma}^{\mathrm{ext}}\) to the rotation about the CM. (In this case you will find there is an extra unknown, the force of friction. You can eliminate this by applying Newton's second law to the motion of the CM. The moment of inertia for rotation about the \(\mathrm{CM}\) is \(\frac{1}{2} M R^{2}\).)
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