/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Two runners of mass \(70 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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Two runners of mass \(70 \mathrm{~kg}\) and \(60 \mathrm{~kg}\), respectively, have a total linear momentum of \(350 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). The heavier runner is running at \(2.0 \mathrm{~m} / \mathrm{s}\). Determine the possible velocities of the lighter runner.

Short Answer

Expert verified
The lighter runner's velocity is 3.5 m/s.

Step by step solution

01

Calculate the Momentum of the Heavier Runner

The linear momentum of an object is the product of its mass and velocity. For the heavier runner with mass 70 kg and velocity 2.0 m/s, the momentum is \( p_1 = m_1 \times v_1 = 70 \times 2.0 = 140 \, \text{kg} \cdot \text{m/s} \).
02

Calculate the Momentum of the Lighter Runner

Since the total momentum of both runners is given as 350 kg·m/s, the momentum of the lighter runner (\( p_2 \)) can be found by subtracting the heavier runner's momentum from the total momentum: \( p_2 = 350 - 140 = 210 \, \text{kg} \cdot \text{m/s} \).
03

Determine the Velocity of the Lighter Runner

The momentum of the lighter runner is the product of its mass and velocity: \( p_2 = m_2 \times v_2 \). With \( m_2 = 60 \, \text{kg} \) and \( p_2 = 210 \, \text{kg} \cdot \text{m/s} \), the velocity \( v_2 \) can be found using the equation: \( v_2 = \frac{p_2}{m_2} = \frac{210}{60} = 3.5 \, \text{m/s} \).

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Key Concepts

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

Physics Problem Solving
When tackling physics problems, there's a simple approach that can make even the most complex problems seem manageable. Begin by identifying all given information and what you need to find out.
Next, write down relevant formulas that might be useful in the context of the problem. This organizes your thoughts and helps determine which concepts to apply.
For instance, in this exercise, we have two runners with known masses and total momentum. We need to identify one runner's unknown velocity. Knowing to use the momentum formula gives our problem-solving a clear direction.

Divide the problem into smaller parts and solve each step systematically. Calculating the momentum of each runner separately and then combining information makes understanding easier. Always check your final results to ensure they make sense contextually. Remember, consistency in units and double-checking calculations are equally as important.
Conservation of Momentum
The conservation of momentum is a crucial principle in physics that states that the total momentum of a closed system remains constant, provided no external forces are acting on it.
In simpler terms, the momentum before and after an event remains the same.
This principle applies to our problem of the two runners. Their combined momentum was given, and we had to use this constant total to find the missing velocity of the lighter runner.
  • Total momentum is simply the sum of individual momenta: \(p_1 + p_2 = ext{constant}\).
  • Changes in momentum are often due to alterations in speed or mass, but with constant mass and unknown velocity, it's clear which variable must be calculated.
Understanding this principle not only helps solve specific problems but also provides a broader understanding of how systems behave in isolated situations.
Momentum Calculation
Momentum characterizes how much motion an object has and is calculated by multiplying mass by velocity.This simple formula - \(p = m \times v\) - is a fundamental concept utilized in many physics applications.
In our exercise with two runners, knowing one runner's mass and velocity allowed us to calculate the individual momentum easily. For the second runner, the formula adapted as:
  • The total momentum known: \(350 \, \text{kg} \cdot \text{m/s}\)
  • Subtracting the heavier runner's momentum: \(140 \, \text{kg} \cdot \text{m/s}\)
  • Leaving lighter runner's momentum as: \(210 \, \text{kg} \cdot \text{m/s}\)
Ultimately solving for the lighter runner's velocity with \(v_2 = \frac{p_2}{m_2} = \frac{210}{60} = 3.5 \, \text{m/s}\).
Clear systematic breakdowns like this demystify the calculations and place emphasis on comprehending rather than memorizing formulas.

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Most popular questions from this chapter

A ball of mass \(200 \mathrm{~g}\) is released from rest at a height of \(2.00 \mathrm{~m}\) above the floor and it rebounds straight up to a height of \(0.900 \mathrm{~m}\). (a) Determine the ball's change in momentum due to its contact with the floor. (b) If the contact time with the floor was \(0.0950 \mathrm{~s}\), what was the average force the floor exerted on the ball, and in what direction?

A \(1200-\mathrm{kg}\) car moving to the right with a speed of \(25 \mathrm{~m} / \mathrm{s}\) collides with a \(1500-\mathrm{kg}\) truck and locks bumpers with the truck. Calculate the velocity of the combination after the collision if the truck is initially (a) at rest, (b) moving to the right with a speed of \(20 \mathrm{~m} / \mathrm{s},\) and (c) moving to the left with a speed of \(20 \mathrm{~m} / \mathrm{s}\).

A major league catcher catches a fastball moving at \(95.0 \mathrm{mi} / \mathrm{h}\) and his hand and glove recoil \(10.0 \mathrm{~cm}\) in bringing the ball to rest. If it took 0.00470 s to bring the ball (with a mass of \(250 \mathrm{~g}\) ) to rest in the glove, (a) what are the magnitude and direction of the change in momentum of the ball? (b) Find the average force the ball exerts on the hand and glove.

At a basketball game, a 120 -lb cheerleader is tossed vertically upward with a speed of \(4.50 \mathrm{~m} / \mathrm{s}\) by a male cheerleader. (a) What is the cheerleader's change in momentum from the time she is released to just before being caught if she is caught at the height at which she was released? (b) Would there be any difference if she were caught \(0.30 \mathrm{~m}\) below the point of release? If so, what is the change then?

Two ice skaters not paying attention collide in a completely inelastic collision. Prior to the collision, skater 1 , with a mass of \(60 \mathrm{~kg},\) has a velocity of \(5.0 \mathrm{~km} / \mathrm{h}\) eastward, and moves at a right angle to skater \(2,\) who has a mass of \(75 \mathrm{~kg}\) and a velocity of \(7.5 \mathrm{~km} / \mathrm{h}\) southward. What is the velocity of the skaters after collision?

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