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Chapter 9: Problem 94

An old Chrysler with mass \(2400 \mathrm{~kg}\) is moving along a straight stretch of road at \(80 \mathrm{~km} / \mathrm{h}\). It is followed by a Ford with mass 1600 kg moving at \(60 \mathrm{~km} / \mathrm{h}\). How fast is the center of mass of the two cars moving?

Short Answer

Expert verified
The center of mass of the two cars is moving at 20 m/s.

Step by step solution

01

Convert velocities to meters per second

We need to convert the velocities from km/h to m/s. To do this, use the conversion factor: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\).Chrysler's velocity: \(80 \text{ km/h} \times \frac{1}{3.6} \text{ m/s} = 22.22 \text{ m/s}\).Ford's velocity:\(60 \text{ km/h} \times \frac{1}{3.6} \text{ m/s} = 16.67 \text{ m/s}\).
02

Write the formula for center of mass velocity

The velocity of the center of mass \(v_{cm}\) of a system of particles is given by:\[ v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \]where \(m_1\) and \(v_1\) are the mass and velocity of the first object (Chrysler), and \(m_2\) and \(v_2\) are the mass and velocity of the second object (Ford).
03

Substitute known values into the equation

Substitute the given masses and velocities into the formula:\[ v_{cm} = \frac{(2400 \text{ kg} \times 22.22 \text{ m/s}) + (1600 \text{ kg} \times 16.67 \text{ m/s})}{2400 \text{ kg} + 1600 \text{ kg}} \]
04

Calculate the components of the numerator

Calculate the momentum for each car:\(2400 \text{ kg} \times 22.22 \text{ m/s} = 53328 \text{ kg} \cdot \text{m/s}\).\(1600 \text{ kg} \times 16.67 \text{ m/s} = 26672 \text{ kg} \cdot \text{m/s}\).
05

Sum the components and divide by total mass

Add the two products and divide by the total mass:\[ v_{cm} = \frac{53328 + 26672}{4000} \]This simplifies to: \[ v_{cm} = \frac{80000}{4000} = 20 \text{ m/s} \].

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Key Concepts

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

Velocity Conversion
When dealing with physics problems, often you'll need to change units to make calculations easier. Like when velocities are in kilometers per hour, but your equations require meters per second. A simple way to do this is using the conversion factor:
  • 1 km/h = \( \frac{1}{3.6} \text{ m/s} \).
Convert each car's speed systematically. For example:
  • The Chrysler moves at 80 km/h. Multiply 80 by \( \frac{1}{3.6} \) to get 22.22 m/s.
  • The Ford moves at 60 km/h. Multiply 60 by \( \frac{1}{3.6} \) to get 16.67 m/s.
These steps ensure you correctly prepare velocities for further calculations in physics problems. Calculating in consistent units helps keep equations straightforward.
Momentum Calculation
Momentum, a vital concept in physics, measures the movement quantity of an object. It's the product of mass and velocity and is expressed in kilogram meters per second (kg路m/s). The formula is:
  • Momentum = Mass 脳 Velocity
To solve such problems, calculate the momentum of each object separately:
  • For the Chrysler: Multiply its mass, 2400 kg, by its velocity, 22.22 m/s. This gives 53328 kg路m/s.
  • For the Ford: Multiply its mass, 1600 kg, by its velocity, 16.67 m/s. This results in 26672 kg路m/s.
Calculating the momentum of all objects in your physics problem is critical for analyzing systems, such as finding the center of mass velocity.
Mass and Velocity
Mass and velocity form the backbone of many dynamics computations, often helping to determine object behaviors in motion. This is crucial when calculating another concept, the center of mass velocity.
When two objects are involved, find each one's contribution by their mass and modified velocity. This relationship emphasizes how mass and speed affect a system's overall dynamics.
In this scenario, the center of mass velocity formula is:
  • \( v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \).
For our example:
  • Combine 2400 kg and 22.22 m/s for the Chrysler with 1600 kg and 16.67 m/s for the Ford.
  • This results in a balanced view of how combined mass and velocities determine the movement of both cars as a single system.
Grasping the interplay between mass and velocity is crucial in addressing complex physics problems where more than one object interacts.
Physics Problem Solving
Physics problem-solving begins with the identification of known elements and required outcomes. Breaking down a problem into smaller logical steps is vital.
To address the exercise regarding the moving cars, start by converting velocities into a consistent unit (like m/s). With converted values, apply the center of mass velocity formula:
  • Identify all formula components: the masses and velocities of each object involved.
  • Plug these into the equation \( v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \).
Next, compute individual components, like total momentum.
  • Add results for a final numerator before division by the total mass.
  • The solution offers how fast the two-car system's center of mass moves at 20 m/s.
This systematic approach helps dissect any complex physics problem, ensuring clarity and accuracy in your results.

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Most popular questions from this chapter

Two particles \(P\) and \(Q\) are released from rest \(1.0 \mathrm{~m}\) apart. \(P\) has a mass of \(0.10 \mathrm{~kg}\), and \(Q\) a mass of \(0.30 \mathrm{~kg} . P\) and \(Q\) attract each other with a constant force of \(1.0 \times 10^{-2} \mathrm{~N}\). No external forces act on the system. (a) What is the speed of the center of mass of \(P\) and \(Q\) when the separation is \(0.50 \mathrm{~m} ?(\mathrm{~b})\) At what distance from \(P\) 's original position do the particles collide?

In Fig. 9-59, a \(10 \mathrm{~g}\) bullet moving directly upward at 1000 \(\mathrm{m} / \mathrm{s}\) strikes and passes through the center of mass of a \(5.0 \mathrm{~kg}\) block initially at rest. The bullet emerges from the block moving directly upward at \(400 \mathrm{~m} / \mathrm{s}\). To what maximum height does the block then rise above its initial position?

At time \(t=0\), force \(\vec{F}_{1}=(-4.00 \hat{\mathrm{i}}+5.00 \hat{\mathrm{j}}) \mathrm{N}\) acts on an initially stationary particle of mass \(2.00 \times 10^{-3} \mathrm{~kg}\) and force \(\vec{F}_{2}=(2.00 \hat{\mathrm{i}}-4.00 \hat{\mathrm{j}}) \mathrm{N}\) acts on an initially stationary particle of mass \(4.00 \times 10^{-3} \mathrm{~kg}\). From time \(t=0\) to \(t=2.00 \mathrm{~ms}\), what are the (a) magnitude and (b) angle (relative to the positive direction of the \(x\) axis) of the displacement of the center of mass of the twoparticle system? (c) What is the kinetic energy of the center of mass at \(t=2.00 \mathrm{~ms} ?\)

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?

Particle \(A\) and particle \(B\) are held together with a compressed spring between them. When they are released, the spring pushes them apart, and they then fly off in opposite directions, free of the spring. The mass of \(A\) is \(2.00\) times the mass of \(B\), and the energy stored in the spring was \(60 \mathrm{~J}\). Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energies of (a) particle \(A\) and (b) particle \(B\) ?
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