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Chapter 8: Problem 93

You are standing on a concrete slab that in turn is resting on a frozen lake. Assume there is no friction between the slab and the ice. The slab has a weight five times your weight. If you begin walking forward at 2.00 m/s relative to the ice, with what speed, relative to the ice, does the slab move?

Short Answer

Expert verified
The slab moves at 0.40 m/s in the opposite direction.

Step by step solution

01

Define the System

Consider the system consisting of you and the slab resting on the ice. Since there is no external horizontal force (like friction), the conservation of momentum applies to the system.
02

Set Up the Conservation of Momentum Equation

Initially, both you and the slab are stationary relative to the ice. Therefore, the initial momentum of the system is zero. When you start walking with a speed of 2.00 m/s relative to the ice, the slab must move in the opposite direction to conserve momentum. Let your mass be \( m \) and the slab's mass be \( 5m \). With your walking speed as \( v = 2.00 \text{ m/s} \), set \( V \) as the slab's speed relative to the ice.
03

Apply Momentum Conservation

According to momentum conservation: \[ m v + 5m V = 0 \]This equation represents that the total momentum remains zero, where \( m v \) is your momentum and \( 5m V \) is the slab's opposing momentum.
04

Solve for the Slab's Velocity

Rearrange the equation to find \( V \):\[ 5m V = -m v \]\[ V = -\frac{m v}{5m} = -\frac{v}{5} \]Substitute \( v = 2.00 \text{ m/s} \):\[ V = -\frac{2.00}{5} = -0.40 \text{ m/s} \]The negative sign indicates the slab moves in the opposite direction of your walking.

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Key Concepts

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

Physics Problem Solving
Physics problem solving can often feel like solving a puzzle. The key is to break down the problem into smaller, manageable parts. Start by carefully reading the problem to understand what is given and what is being asked.
In physics, drawing diagrams can be very helpful to visualize the scenario. Next, identify the relevant physics concepts involved.
In our case, the problem involves momentum conservation and the motion on a frictionless surface.
  • Understand the initial conditions: you and the slab are at rest.
  • Note the actions: you start walking.
  • Identify the result: the slab's movement.
Taking a systematic approach helps to ensure you consider all parts of the problem, preventing oversight of important details.
Frictionless Surface
A frictionless surface is an idealized surface where no frictional force opposes the motion of objects. This means that once an object starts moving, it will continue moving indefinitely on such a surface unless acted upon by another force.
In our problem, the ice is assumed to be a frictionless surface. This assumption simplifies the problem by eliminating external horizontal forces.
This lack of friction is crucial for applying the conservation of momentum, as no momentum is lost or gained from external forces. Real-world surfaces are never truly frictionless, but this assumption helps focus on the internal dynamics of the system.
Relative Velocity
Relative velocity refers to the velocity of one object as observed from another moving object or reference frame.
In the problem, your velocity is given as 2.00 m/s relative to the ice. Meanwhile, the slab moves with a certain velocity in the opposite direction, also relative to the ice.
  • You are moving relative to the ice.
  • The slab reacts to your motion.
  • The velocities are compared with a common reference: the ice.
Understanding relative velocity is essential for calculating how different parts of the system interact and influence each other's motion.
Conservation Laws in Physics
Conservation laws are foundational principles in physics, positing that certain quantities remain constant in a closed system. One of these laws is the conservation of momentum.
According to this law, the total momentum of a closed system remains constant if no external forces act upon it.
  • Total initial momentum = Total final momentum.
  • In our exercise, the system is closed because no external horizontal forces influence it.
  • This allows us to equate the momentum before and after you start walking.
By applying the conservation of momentum, we were able to find the speed at which the slab moves in response to your walking.

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Most popular questions from this chapter

Two cars collide at an intersection. Car \(A\), with a mass of 2000 kg, is going from west to east, while car \(B\), of mass 1500 kg, is going from north to south at 15 m/s. As a result, the two cars become enmeshed and move as one. As an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65\(^\circ\) south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car \(A\) going just before the collision?

\(\textbf{Combining Conservation Laws}\). A 5.00-kg chunk of ice is sliding at 12.0 m/s on the floor of an ice-covered valley when it collides with and sticks to another 5.00-kg chunk of ice that is initially at rest \((\textbf{Fig. P8.73})\). Since the valley is icy, there is no friction. After the collision, how high above the valley floor will the combined chunks go?

An astronaut in space cannot use a conventional means, such as a scale or balance, to determine the mass of an object. But she does have devices to measure distance and time accurately. She knows her own mass is 78.4 kg, but she is unsure of the mass of a large gas canister in the airless rocket. When this canister is approaching her at 3.50 m/s, she pushes against it, which slows it down to 1.20 m/s (but does not reverse it) and gives her a speed of 2.40 m/s. What is the mass of this canister?

A 12.0-kg shell is launched at an angle of 55.0\(^\circ\) above the horizontal with an initial speed of 150 m/s. At its highest point, the shell explodes into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Ignore air resistance. If the heavier fragment lands back at the point from which the shell was launched, where will the lighter fragment land, and how much energy was released in the explosion?

A 0.0450-kg golf ball initially at rest is given a speed of 25.0 m/s when a club strikes it. If the club and ball are in contact for 2.00 ms, what average force acts on the ball? Is the effect of the ball's weight during the time of contact significant? Why or why not?
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