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At in preparation for this problem. Two friends, Al and Jo, have a combined mass of \(168 \mathrm{~kg}\). At an ice skating rink they stand close together on skates, at rest and facing each other, with a compressed spring between them. The spring is kept from pushing them apart because they are holding each other. When they release their arms, Al moves off in one direction at a speed of \(0.90 \mathrm{~m} / \mathrm{s}\), while Jo moves off in the opposite direction at a speed of \(1.2 \mathrm{~m} / \mathrm{s}\). Assuming that friction is negligible, find Al's mass.

Short Answer

Expert verified
Al's mass is 96 kg.

Step by step solution

01

Understand the Concept of Conservation of Momentum

The principle of conservation of momentum states that in the absence of external forces, the total momentum of a closed system remains constant. Here, Al and Jo are initially at rest, so the total initial momentum is zero. When they move apart, their momentum should still sum to zero.
02

Set Up the Equation Using Conservation of Momentum

Let Al's mass be \(m_A\) and Jo's mass be \(m_J\). The total initial momentum is 0, and the final momentum is the sum of the momenta of both friends. Thus, we write: \[m_A \times 0.90 + m_J \times (-1.2) = 0.\] We have the opposite signs because they move in opposite directions.
03

Express Jo’s Mass in Terms of Al's Mass

Since their combined mass is given as \(m_A + m_J = 168\), we can express Jo's mass as \(m_J = 168 - m_A\). Substituting this into the momentum equation gives: \[m_A \times 0.90 + (168 - m_A) \times (-1.2) = 0.\]
04

Simplify and Solve the Equation for Al's Mass

Substitute \(m_J = 168 - m_A\) in the momentum equation and simplify: \[0.90m_A - 1.2(168 - m_A) = 0\]. Distribute the \(-1.2\): \[0.90m_A - 201.6 + 1.2m_A = 0.\] Combine like terms: \[2.1m_A - 201.6 = 0.\] Add 201.6 to both sides: \[2.1m_A = 201.6.\] Divide by 2.1: \[m_A = \frac{201.6}{2.1} = 96.\]
05

Verify the Solution

Now, verify Al's mass by checking Jo's calculated mass: \(m_J = 168 - m_A = 168 - 96 = 72\). Check total momentum: \(96 \times 0.90 = 86.4\) and \(72 \times 1.2 = 86.4\). These are equal and opposite, confirming the conservation of momentum.

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Key Concepts

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

Understanding Ice Skating Physics
Ice skating physics is a fascinating subject that captures the laws of motion on the slippery surface of ice. One of the key concepts in ice skating is friction, or rather the lack of it, which allows skaters to glide smoothly across the ice. The surface of the ice offers minimal resistance, enabling movements that would otherwise be challenging on rougher surfaces.

In scenarios like this exercise, where two skaters push off each other, another important physics concept is brought into play: the conservation of momentum. When Al and Jo stand on the ice with a spring between them, they create a closed system. When they release their arms, despite the minimal friction, the laws of physics ensure that they move in predictable ways. Because they are initially at rest and then push off each other with equal force, their velocities are determined by their respective masses and the conservation of momentum.

Ultimately, understanding ice skating physics helps us appreciate how skaters skillfully utilize these physical principles to maneuver gracefully or interact with others on the ice.
Mastering Mass Calculation
Calculating the mass of objects in motion, such as skaters on ice, involves using the principles of physics to draw relationships between weight and motion. In our exercise, we are tasked with finding Al's mass, using the given combined mass of Al and Jo, which is 168 kg.

Step one is to establish formulas for the masses. For instance, if we know the total mass—their combined mass—we can express Jo’s mass in terms of Al’s. The relationship can be described as:
  • Jo's mass, denoted as \(m_J\), can be expressed as \(m_J = 168 - m_A\) if \(m_A\) is Al's mass.
This relationship is crucial to solve for unknown variables within the equation of momentum, providing a path to understand the interactions between their velocities and weights.

Mastering mass calculations requires attention to detail and understanding the interplay between different variables, making it a critical skill in solving physics problems.
Momentum Equation Insights
The momentum equation is a powerful tool in physics that helps us predict motion outcomes in systems, especially in contexts with no external forces, like our ice skating example. The conservation of momentum formula is given by the sum of initial momentums equalling the sum of final momentums of a system.

For Al and Jo, initially at rest, their combined momentum is zero. When they move apart on releasing their hold on each other, their momentums are equal and opposite. By letting Al's mass be \(m_A\) and assuming Jo's mass is \(m_J = 168 - m_A\), we set up the following momentum conservation equation:
  • \(m_A \times 0.90 + m_J \times (-1.2) = 0\).
Breaking it down, the velocities are multiplied by their corresponding masses, producing motion equal in magnitude but opposite in direction.

By substituting \(m_J = 168 - m_A\) into the equation and solving, we find \(m_A = 96\). Checking our result confirms the balance of momenta, exemplifying the power and reliability of momentum equations in everyday physics applications.

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Most popular questions from this chapter

Multiple-Concept Example 7 deals with some of the concepts that are used to solve this problem. A cue ball (mass \(=0.165 \mathrm{~kg}\) ) is at rest on a frictionless pool table. The ball is hit dead center by a pool stick, which applies an impulse of \(+1.50 \mathrm{~N}\) s to the ball. The ball then slides along the table and makes an elastic head-on collision with a second ball of equal mass that is initially at rest. Find the velocity of the second ball just after it is struck.

A two-stage rocket moves in space at a constant velocity of \(4900 \mathrm{~m} / \mathrm{s} .\) The two stages are then separated by a small explosive charge placed between them. Immediately after the explosion the velocity of the 1200 -kg upper stage is \(5700 \mathrm{~m} / \mathrm{s}\) in the same direction as before the explosion. What is the velocity (magnitude and direction) of the 2400 -kg lower stage after the explosion?

ssm Two people are standing on a \(2.0\) -m-long platform, one at each end. The platform floats parallel to the ground on a cushion of air, like a hovercraft. One person throws a \(6.0\) -kg ball to the other, who catches it. The ball travels nearly horizontally. Excluding the ball, the total mass of the platform and people is \(118 \mathrm{~kg}\). Because of the throw, this 118 -kg mass recoils. How far does it move before coming to rest again?

Two balls are approaching each other head-on. Their velocities are +9.70 and \(-11.8 \mathrm{~m} /\) s. Determine the velocity of the center of mass of the two balls (a) if they have the same mass and (b) if the mass of one ball \((v=9.70 \mathrm{~m} / \mathrm{s})\) is twice the mass of the other ball \((v=-11.8 \mathrm{~m} / \mathrm{s})\).

Part \(a\) of the drawing shows a bullet approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part \(b\). Note that both blocks are moving after the collision with the bullet. (a) Can the conservation of linear momentum be applied to this three-object system, even though the second collision occurs a bit later than the first one? Justify your answer. Neglect any mass removed from the first block by the bullet. (b) Is the total kinetic energy of this three-body system conserved? If not, would the total kinetic energy after the collisions be greater than or smaller than that before the collisions? Justify your answer.

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