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A \(91 \mathrm{~kg}\) man lying on a surface of negligible friction shoves a \(68 \mathrm{~g}\) stone away from himself, giving it a speed of \(4.0 \mathrm{~m} / \mathrm{s}\). What speed does the man acquire as a result?

Short Answer

Expert verified
The man's speed is approximately \(-0.003 \mathrm{~m/s}\), opposite to the stone's direction.

Step by step solution

01

Understanding the Problem

We need to calculate the speed acquired by a man when he pushes a stone away from himself. This is a conservation of momentum problem where the total momentum before and after the action remains the same, due to the principle of conservation of linear momentum in isolated systems.
02

Identifying Given Values

The mass of the man, \( m_1 \), is \( 91 \mathrm{~kg} \). The mass of the stone, \( m_2 \), is \( 68 \mathrm{~g} \), which is \( 0.068 \mathrm{~kg} \). The velocity of the stone after being shoved, \( v_2 \), is \( 4.0 \mathrm{~m/s} \). The initial velocities of both the man and the stone are \( 0 \mathrm{~m/s} \). We need to find the velocity of the man, \( v_1 \), after the shove.
03

Applying Conservation of Momentum

Since no external forces act on the system (the man and stone), the total momentum before and after shoving must be equal. Therefore, we use the formula: \[ m_1 \, v_1 + m_2 \, v_2 = 0. \]
04

Solving for the Man's Velocity

Rearrange the equation to solve for \( v_1 \): \[ v_1 = - \frac{m_2 \, v_2}{m_1}. \]
05

Substituting Known Values

Substitute the known values into the equation: \[ v_1 = - \frac{0.068 \, \mathrm{kg} \times 4.0 \mathrm{~m/s}}{91 \, \mathrm{kg}} \approx -0.003 \mathrm{~m/s}. \]
06

Calculating the Result

Perform the calculation: \[ v_1 \approx -0.003 \mathrm{~m/s}. \] The negative sign indicates that the man's velocity is in the opposite direction to the stone's velocity.

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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 is about breaking down complex situations into small manageable parts. Problems like the one involving the man and the stone can initially seem confusing. However, by identifying given values, applying formulas, and executing calculations step-by-step, you can find the solution.
The majority of physics problems require this systematic approach:
  • Understand the problem: Break it into parts and recognize which principles apply.
  • Identify known and unknown values: List them clearly to have a structured view.
  • Apply relevant formulas: Use the information you gathered to set up equations.
  • Solve step-by-step: Isolate variables and perform calculations carefully.
For instance, in the exercise above, conservation of momentum is the key principle. Knowing this, you can confidently proceed through the steps outlined.
Solving physics problems is like following a map, where each step leads you a bit closer to your destination.
Momentum
Momentum is a fundamental concept in physics. It describes the quantity of motion an object has, which depends both on its mass and its velocity. Simple and powerful, momentum can be expressed through the formula: \[ p = m imes v \] where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity.
Momentum is a vector quantity, which means it has both magnitude and direction. This directional aspect is crucial, particularly in problems involving two bodies interacting, like the man and the stone. When the stone is shoved, both bodies exert forces on each other in opposite directions.
Momentum helps explain these interactions easily: the momentum gained by the stone is equal in magnitude, but opposite in direction, to the momentum gained by the man.
This principle helps solve many real-world physics problems, providing a tool to predict how objects will move and interact. Understanding momentum builds a foundation for tackling more complex concepts in physics.
Linear Momentum
Linear momentum, often simply called momentum, pertains specifically to motion along a straight line. This is the case in our exercise, where the man and the stone move in a straight line away from each other after the shove.
Because momentum is conserved in isolated systems, linear momentum tells us that the total momentum before and after an interaction remains constant.
  • In systems like the man and stone, no external forces mean we can use conservation rules.
  • The total initial momentum (which is zero since neither the man nor stone is moving) must equal the total final momentum.
  • By setting up the equation \( m_1 \, v_1 + m_2 \, v_2 = 0 \), we can find unknown velocities by solving for the variables.
Linear momentum offers a streamlined way to solve complex interactions, making it a cornerpiece of classical mechanics.

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Most popular questions from this chapter

A man (weighing \(915 \mathrm{~N}\) ) stands on a long railroad flatcar (weighing \(2415 \mathrm{~N}\) ) as it rolls at \(18.2 \mathrm{~m} / \mathrm{s}\) in the positive direction of an \(x\) axis, with negligible friction. Then the man runs along the flatcar in the negative \(x\) direction at \(4.00 \mathrm{~m} / \mathrm{s}\) relative to the flatcar. What is the resulting increase in the speed of the flatcar?

Block \(1,\) with mass \(m_{1}\) and speed \(4.0 \mathrm{~m} / \mathrm{s},\) slides along an \(x\) axis on a frictionless floor and then undergoes a one-dimensional elastic collision with stationary block \(2,\) with mass \(m_{2}=0.40 m_{1}\). The two blocks then slide into a region where the coefficient of kinetic friction is 0.50 ; there they stop. How far into that region do (a) block 1 and (b) block 2 slide?

block 1 of mass \(m_{1}\) slides along an \(x\) axis on a frictionless floor with a speed of \(v_{1 j}=4.00 \mathrm{~m} / \mathrm{s}\). Then it undergoes a one- dimensional elastic collision with stationary block 2 of mass \(m_{2}=0.500 m_{1}\). Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass \(m_{3}=0.500 \mathrm{~m}_{2}\). (a) What then is the speed of block 3 ? Are (b) the speed, (c) the kinetic energy, and (d) the momentum of block 3 greater than, less than, or the same as the initial values for block \(1 ?\)

A ball having a mass of \(150 \mathrm{~g}\) strikes a wall with a speed of \(5.2 \mathrm{~m} / \mathrm{s}\) and rebounds with only \(50 \%\) of its initial kinetic energy. (a) What is the speed of the ball immediately after rebounding? (b) What is the magnitude of the impulse on the wall from the ball? (c) If the ball is in contact with the wall for \(7.6 \mathrm{~ms}\), what is the magnitude of the average force on the ball from the wall during this time interval?

A rocket is moving away from the solar system at a speed of \(6.0 \times 10^{3} \mathrm{~m} / \mathrm{s} .\) It fires its engine, which ejects exhaust with a speed of \(3.0 \times 10^{3} \mathrm{~m} / \mathrm{s}\) relative to the rocket. The mass of the rocket at this time is \(4.0 \times 10^{4} \mathrm{~kg},\) and its acceleration is \(2.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the thrust of the engine? (b) At what rate, in kilograms per second, is exhaust ejected during the firing?

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