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Chapter 8: Problem 24

You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock so it has velocity 12.0 \(\mathrm{m} / \mathrm{s}\) relative to the earth at an angle of \(35.0^{\circ}\) above the borizontal. If your mass is 70.0 \(\mathrm{kg}\) and the rock's mass is 15.0 kg, what is your speed after you throw the rock (see Discussion Question \(Q 8.7\) ?

Short Answer

Expert verified
The person's speed after throwing the rock is 2.10 m/s.

Step by step solution

01

Identify known values

We have the mass of the person, m_person = 70.0 kg, and the mass of the rock, m_rock = 15.0 kg. The velocity of the rock, v_rock, is 12.0 m/s at an angle 35.0掳 above the horizontal. We need to find the person's speed, v_person, after throwing the rock.
02

Apply conservation of momentum

Since there are no external forces acting in the horizontal direction, we can apply the conservation of momentum. The initial momentum is zero because both the person and the rock are stationary. After throwing the rock, the total momentum must still be zero.
03

Calculate horizontal velocity of the rock

The horizontal component of the rock's velocity can be found using trigonometry: \[ v_{rock_x} = v_{rock} \cdot \cos(35.0^{\circ}) \approx 12.0 \cdot 0.819 = 9.828 \text{ m/s}. \]
04

Set up the equation for momentum conservation

Using the conservation of momentum in the horizontal direction: \[ m_{person} \cdot v_{person} + m_{rock} \cdot v_{rock_x} = 0. \] We want to find v_person, so: \[ 70 \cdot v_{person} + 15 \cdot 9.828 = 0. \]
05

Solve for the person's velocity

Rearrange the equation to solve for v_person:\[ v_{person} = -\frac{15 \cdot 9.828}{70} = -2.104. \] The negative sign indicates that the velocity is in the opposite direction to the rock鈥檚 horizontal velocity.

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Key Concepts

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

Frictionless Surface
Imagine standing on a sheet of ice that is perfectly smooth 鈥� it's a frictionless surface. This concept is crucial in physics because it helps us focus only on the forces acting internally without external resistance interfering. The absence of friction means that when any action is taken, like throwing a rock, there are no external horizontal forces.
In our exercise, no friction acts on the person or rock. This condition allows the conservation of momentum to apply directly, simplifying our calculations significantly.
  • Key takeaway: On a frictionless surface, any force applied results solely in internal momentum changes.
  • Result: You fly backward when you throw the rock due to action-reaction pairs, here described by Newton's third law.
Projectile Motion
When you throw an object, like the rock in our example, it follows a path known as projectile motion. Projectile motion encompasses both horizontal and vertical motion, and it originates at the object's launch.
For this particular exercise, we needed to calculate only the horizontal component of the rock's velocity. Using the angle of throw, 35 degrees, we determined the horizontal velocity to be approximately 9.828 m/s.
You might be wondering why just the horizontal component matters. It's because the conservation of momentum is considered separately in different directions.
  • Key aspect: The angle of the launch affects both components, but in horizontal momentum conservation, we focus on the horizontal aspect.
  • Formula snapshot: Horizontal velocity is determined using \( v_{rock_x} = v_{rock} \cdot \cos(35.0^{\circ}) \).
Physics Problem Solving
Solving physics problems can seem daunting, but it becomes manageable when approached systematically. Imagine it as a four-step strategy:
  • Identifying known values: Determine what you immediately know 鈥� such as masses and velocities.
  • Applying principles: Usage of key physics laws or principles, like the conservation of momentum in this example.
  • Breaking down calculations: Use mathematical tools, such as trigonometry, to break complex movements into simpler components.
  • Solving for unknowns: Rearrange your equations to find the desired unknowns, here exemplified by finding your velocity after throwing the rock.
In this exercise, following such structured steps leads us directly to the correct solution. Remember, an equation is just a mathematical statement of a physical reality, so understanding it conceptually helps you navigate problems with ease.

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Most popular questions from this chapter

In a volcanic eruption, a 2400 -kg boulder is thrown vertically upward into the air. At its highest point, it suddenly explodes (due to trapped gases) into two fragments, one being three times the mass of the other. The lighter fragment starts out with only horizontal velocity and lands 274 \(\mathrm{m}\) directly north of the point of the explosion. Where will the other fragment land? Neglect any air resistance.

A radio-controlled model airplane has a momentum given by \(\left[\left(-0.75 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}^{3}\right) t^{2}+\right.\) \((3.0 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}) ] \hat{\mathrm{i}}+\left(0.25 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}^{2}\right) \hat{t}\) What are the \(x-y\) , and z-components of the net force on the airplane?

An \(8.00-\mathrm{kg}\) ball, hanging from the ceiling by a light wire 135 \(\mathrm{cm}\) long, is struck in an elastic collision by a \(2.00-\mathrm{kg}\) ball moving horizontally at 5.00 \(\mathrm{m} / \mathrm{s}\) just before the collision. Find the tension in the wire just after the collision.

In beta decay, a nucleus emits an electron. A 210 Bi (bismuth) nucleus at rest undergoes beta decay to \(^{210} \mathrm{Po}\) (polonium). Suppose the emitted electron moves to the right with a momentum of \(5.60 \times 10^{-22} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) . The 210 \(\mathrm{Po}\) nucleus, with mass \(3.50 \times 10^{-25} \mathrm{kg},\) recoils to the left at a speed of \(1.14 \times 10^{-3} \mathrm{m} / \mathrm{s}\) Momentum conservation requires that a second particle, called an antineutrino, must also be emitted. Calculate the magnitude and direction of the momentum of the antineutrino that is emitted in this decay.

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. \(A 6.50-k g\) squid (including the water in the cavity) at rest suddenly sees a dangerous predator. (a) If the squid has 1.75 \(\mathrm{kg}\) of water in its cavity, at what speed must it expel this water to suddenly achieve a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) to escape the predator? Neglect any drag effects of the surrounding water. (b) How much kinetic energy does the squid create by this maneuver?
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