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Chapter 8: Problem 50

A 1200 -kg station wagon is moving along a straight highway at 12.0 \(\mathrm{m} / \mathrm{s}\) . Another car, with mass 1800 \(\mathrm{kg}\) and speed 20.0 \(\mathrm{m} / \mathrm{s}\) , has its center of mass 40.0 \(\mathrm{m}\) ahead of the center of mass of the station wagon (Fig. 8.39\() .\) (a) Find the position of the center of mass of the system consisting of the two automobiles. (b) Find the magnitude of the total momentum of the system from the given data. (o) Find the speed of the center of mass of the system. (d) Find the total momentum of the system, using the speed of the center of mass. Compare your result with that of part (b).

Short Answer

Expert verified
(a) 24 m, (b) 50400 kg路m/s, (c) 16.8 m/s, (d) 50400 kg路m/s; consistent with (b).

Step by step solution

01

Understanding the Problem

Given two cars, one with mass 1200 kg at 12.0 m/s and another with mass 1800 kg at 20.0 m/s, we need to find the center of mass of the two-car system and related details about momentum.
02

Calculate the Position of the Center of Mass

The position of the center of mass of a system of particles is calculated using the formula \( x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} \), where \( m_1 \) and \( m_2 \) are the masses, and \( x_1 \) and \( x_2 \) are their positions. Assume the station wagon is at position 0: \( x_{cm} = \frac{(1200 \times 0) + (1800 \times 40)}{1200 + 1800} = \frac{72000}{3000} = 24 \) m.
03

Calculate the Total Momentum of the System

The total momentum \( p_{total} \) is the sum of the momenta of all individual objects: \( p_{total} = m_1v_1 + m_2v_2 \). Substitute the values: \( p_{total} = 1200 \times 12 + 1800 \times 20 = 14400 + 36000 = 50400 \) kg路m/s.
04

Calculate the Speed of the Center of Mass

The speed of the center of mass \( v_{cm} \) is the total momentum divided by the total mass: \( v_{cm} = \frac{p_{total}}{m_1 + m_2} = \frac{50400}{3000} = 16.8 \) m/s.
05

Check Momentum with Speed of Center of Mass

Calculate the total momentum using the speed of the center of mass: \( p_{total} = (m_1 + m_2) \times v_{cm} = 3000 \times 16.8 = 50400 \) kg路m/s. This matches the result from Step 2, confirming our calculations.

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Key Concepts

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

Center of Mass Calculation
When dealing with problems involving multiple objects, such as cars, it's essential to determine the center of mass. This point represents where you can consider the entire mass of a system to be concentrated. Calculating the center of mass is straightforward. You use the formula: \[ x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} \]This formula accounts for the masses (\(m_1, m_2\)) and positions (\(x_1, x_2\)) of the objects involved. In our problem, treat the station wagon's position as zero. Hence, \(x_1 = 0\) and \(x_2 = 40\) meters ahead. Substituting the values, we find the center of mass \(x_{cm}\) is 24 meters. This position indicates that the system's combined mass behaves like a single point located at this distance.
Momentum Conservation
Momentum is a crucial concept when analyzing any system of particles. It's a measure of an object's motion and can be calculated by multiplying an object's mass by its velocity. In physics, total momentum is conserved in a closed system unless external forces are present. For the two-car system in this exercise, the total momentum is the sum of their individual momenta: \[ p_{total} = m_1v_1 + m_2v_2 \]This results in \(14400 + 36000 = 50400\) kg路m/s. This total momentum remains unchanged as long as no external forces interfere. It's this conservation principle that allows us to check our calculations' accuracy.
System of Particles
A system of particles, such as two cars moving along a highway, requires handling multiple objects as a single system. Simplifying how we examine them helps focus on core dynamics. Using their combined mass and motion, we can predict their collective behavior. In this context, the total mass of the system is simply the sum of the individual masses: \(m_1 + m_2\). This simplification aids in calculations such as finding the center of mass and determining overall kinetic properties. Given their combined masses, it becomes easier to calculate the system's collective momentum and speed. The system's center of mass and its speed provide valuable insight into how the grouped particles will behave during their motion.
Physics Education
Understanding concepts like center of mass and momentum within a system of particles is fundamental in physics education. These principles allow students to solve real-world problems with accuracy and confidence. Physics is the study of natural phenomena and equips learners with problem-solving skills crucial in daily life. Approach these problems with these simple steps:
  • Identify the system and its components.
  • Calculate key properties like center of mass using provided formulas.
  • Ensure comprehension of conservation laws, such as momentum conservation.
  • Compare calculated results to expected outcomes to confirm understanding.
Developing these foundational skills allows students to tackle more complex topics in physics and engineering confidently.

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Most popular questions from this chapter

A \(20.00-\mathrm{kg}\) lead sphere is hanging from a hook by a thin wire 3.50 \(\mathrm{m}\) long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a \(5.00-\mathrm{kg}\) steel dart that embeds itself in the lead sphere. What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?

At time \(t=0,\) a \(2150-\mathrm{kg}\) rocket in outer space fires an engine that exerts an increasing force on it in the \(+x\) -direction. This force obeys the equation \(F_{x}=A t^{2},\) where \(t\) is time, and has a magnitude of 781.25 \(\mathrm{N}\) when \(t=1.25 \mathrm{s}\) . (a) Find the SI value of the constant \(A,\) including its units. \((\mathrm{b})\) What impulse does the engine exert on the rocket during the 1.50 -s interval starting 2.00 s after the engine is fired? (c) By how much does the rocket's velocity change during this interval?

A rocket is fired in deep space, where gravity is negligible. In the first second it ejects \(\frac{1}{160}\) of its mass as exhaust gas and has an accolcration of 15.0 \(\mathrm{m} / \mathrm{s}^{2} .\) What is the spoced of tho cxhaust gas relative to the rocket?

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass \(70.0 \mathrm{kg},\) is moving to the right at 2.00 \(\mathrm{m} / \mathrm{s}\) , while the other. of mass 65.0 \(\mathrm{kg}\) , is moving to the left at 2.50 \(\mathrm{m} / \mathrm{s}\) . What are the magnitude and direction of the velocity of these skaters just after they collide?

When two hydrogen atoms of mass \(m\) combine to form a diatomic hydrogen molecule \(\left(H_{2}\right),\) the potential energy of the system after they combine is \(-\Delta,\) where \(\Delta\) is a positive quantity called the binding energy of the molecule. (a) Show that in a collision that involves only two hydrogen atoms, it is impossible to form an \(\mathrm{H}_{2}\) molecule because momentum and energy cannot simultaneously be conserved. (Hint: If you can show this to be true in one frame of reference, then it is true in all frames of reference. Can you see why?) (b) An \(\mathrm{H}_{2}\) molecule can be formed in a collision that involves three hydrogen atoms. Suppose that before such a collision, each of the three atoms has speed \(1.00 \times 10^{3} \mathrm{m} / \mathrm{s}\) , and they are approaching at \(120^{\circ}\) angles so that at any instant, the atoms lie at the comers of an equilateral triangle. Find the speeds of the \(\mathrm{H}_{2}\) molecule and of the single hydrogen atom that remains after the collision. The binding energy of \(\mathrm{H}_{2}\) is \(\Delta=7.23 \times 10^{-19} \mathrm{J},\) and the mass of the bindrogen atom is \(1.67 \times 10^{-27} \mathrm{kg}\) .
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