/*! 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 54 A \(1200 \mathrm{~kg}\) SUV is m... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(1200 \mathrm{~kg}\) SUV 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 SUV (Fig. E8.54). Find (a) the position of the center of mass of the system consisting of the two cars; (b) the magnitude of the system's total momentum, by using the given data: (c) the speed of the system's center of mass; (d) the system's total momentum, by using the speed of the center of mass. Compare your result with that of part (b).

Short Answer

Expert verified
The position of the center of mass of the system is calculated in Step 1, the total system momentum in Step 2, the speed of the center of mass in Step 3 and the total momentum is validated using another method in Step 4. These steps together solve the exercise in a detailed and methodical way.

Step by step solution

01

Calculate center of mass

The center of mass formula is given by \[ x_{cm} = \frac{\sum m_{i}x_{i}}{\sum m_{i}} \] where \(x_i\) and \(m_i\) are the position and mass of each car respectively. With SUV at position 0m and another car at 40m, center of mass can be found as \[ x_{cm} = \frac{(1200 \mathrm{kg} \cdot 0 \mathrm{m}) + (1800 \mathrm{kg} \cdot 40 \mathrm{m})}{1200 \mathrm{kg} + 1800 \mathrm{kg}} \]
02

Total Momentum Calculation

Total momentum of the system is simply the sum of the momentum of each car. Momentum is given by mass \(\cdot\) speed. Hence, momentum can be calculated as \[ P_{total} = (m_{SUV} \cdot v_{SUV}) + (m_{car} \cdot v_{car}) = (1200 \mathrm{kg} \cdot 12.0 \mathrm{m/s}) + (1800 \mathrm{kg} \cdot 20.0 \mathrm{m/s}) \]
03

Speed of Center of Mass

Speed of the center of mass is calculated as the total momentum of the system divided by the total mass. Hence, it can be calculated as \[ v_{cm} = \frac{P_{total}}{m_{total}} = \frac{P_{total}}{m_{SUV} + m_{car}} \]
04

Validation by Another Method to Calculate Momentum

As a validation step, we should calculate the total momentum using the approach of speed of the center of mass multiplied by the total mass of the system. Hence, \[ P'_{total} = m_{total} \cdot v_{cm} \]. And compare P'_{total} with P_{total} from Step 2. If these two are equal, then all calculations are correct.

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Key Concepts

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

Center of Mass Calculation
In physics, understanding the center of mass is crucial for analyzing the motion of a system with multiple objects. The center of mass of a system is akin to the point where you could balance the entire system if you could place it on a pivot. To find this point mathematically, the formula is \[ x_{cm} = \frac{\sum m_{i}x_{i}}{\sum m_{i}} \]where \(x_i\) represents the position of each object in the system and \(m_i\) represents the mass. In our example, the SUV and the car form a system. The SUV is at a reference point of 0 meters, and the car, 40 meters ahead.
  • SUV: \( m_{SUV} = 1200\, \text{kg},\ x_{SUV} = 0\, \text{m}\)
  • Car: \( m_{car} = 1800\, \text{kg},\ x_{car} = 40\, \text{m}\)
By applying these values in the center of mass equation, we identify the center of mass of the two vehicles relative to the SUV's reference position of 0 meters. This point represents where the collective system appears to concentrate its mass.
Momentum Calculation
Momentum is a fundamental concept in physics because it describes the quantity of motion an object has. It is calculated as the product of an object's mass and its velocity, expressed in the equation \[ p = m \cdot v \] where \(p\) is the momentum, \(m\) is the mass, and \(v\) is the velocity of the object.With the SUV and the car, the separate momenta are:
  • SUV: \( m_{SUV} = 1200\, \text{kg},\ v_{SUV} = 12.0\, \text{m/s}\)
  • Car: \( m_{car} = 1800\, \text{kg},\ v_{car} = 20.0\, \text{m/s}\)
The total momentum of the system is the sum of each vehicle's momentum. Adding these gives us the total momentum \( P_{total} \) considering both the SUV and the car.Measuring the momentum of a system helps predict how that system behaves when interacting with other systems, such as during collisions.
Physics Problem Solving
Physics problems, especially those involving more than one object like in our scenario, require a systematic approach to solve. Generally, tackling such problems involves the following steps:
  • Defining what is known and identifying what is needed.
  • Applying the relevant physics formulas, such as those for center of mass or momentum.
  • Solving algebraically to isolate the expansion of unknowns using given values.
  • Checking the consistency of derived results with physical laws or different methods.
In our example, alternate method solutions such as computing total momentum by multiplying the total mass by the speed of the center of mass help verify the accuracy of our calculations. This practice underscores the importance of validating results in physics problem-solving to ensure they adhere to theoretical expectations and real-world physics principles.
Speed of Center of Mass
The speed of the center of mass is an important characteristic that provides insight into the overall motion of a system. It is determined by dividing the total momentum of the system by the total mass, as given by this formula: \[ v_{cm} = \frac{P_{total}}{m_{total}} \].This calculation gives a single velocity value that can be thought of as the speed at which the center of mass of the system moves. For our system involving two cars, this concept helps explain how they collectively behave as one unit when their separate motions are combined into a single motion.Knowing this speed is especially valuable in analyzing collisions and momentum conservation because it simplifies complex motion into a single trajectory that can be easier to work with for predicting future states of motion.

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Most popular questions from this chapter

\(A 45.0 \mathrm{~kg}\) woman stands up in a \(60.0 \mathrm{~kg}\) canoe \(5.00 \mathrm{~m}\) long. She walks from a point \(1.00 \mathrm{~m}\) from one end to a point \(1.00 \mathrm{~m}\) from the other end (Fig. \(\mathrm{P} 8.92\) ). If you ignore resistance to motion of the canoe in the water, how far does the canoe move during this process?

Object \(B\) is at rest when object \(A\) collides with it. The collision is one- dimensional and elastic. After the collision object \(B\) has half the velocity that object \(A\) had before the collision. (a) Which object has the greater mass? (b) How much greater? (c) If the velocity of object \(A\) before the collision was \(6.0 \mathrm{~m} / \mathrm{s}\) to the right. what is its velocity after the collision?

A Ricocheting Bullet. A \(0.100 \mathrm{~kg}\) stone rests on a frictionless, horizontal surface. A bullet of mass \(6.00 \mathrm{~g}\). traveling horizontally at \(350 \mathrm{~m} / \mathrm{s}\), strikes the stone and rebounds horizontally at right angles to its original direction with a speed of \(250 \mathrm{~m} / \mathrm{s}\). (a) Compute the magnitude and direction of the velocity of the stonc after it is struck. (b) Is the collision perfectly clastic?

Just before it is struck by a racket, a tennis ball weighing \(0.560 \mathrm{~N}\) has a velocity of \((20.0 \mathrm{~m} / \mathrm{s}) \hat{\imath}-(4.0 \mathrm{~m} / \mathrm{s}) \hat{\jmath} .\) During the \(3.00 \mathrm{~ms}\) that the racket and ball are in contact, the net external force on the ball is constant and equal to \(-(380 \mathrm{~N}) \hat{\imath}+(110 \mathrm{~N}) \hat{\jmath}\). What are the \(x\) - and \(y\) -components (a) of the impulse of the net external force applied to the ball; (b) of the final velocity of the ball?

A small wooden block with mass \(0.800 \mathrm{~kg}\) is suspended from the lower end of a light cord that is \(1.60 \mathrm{~m}\) long. The block is initially at rest. A bullet with mass \(12.0 \mathrm{~g}\) is fired at the block with a horizontal velocity \(c_{0}\) - The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of \(0.800 \mathrm{~m}\), the tension in the cord is \(4.80 \mathrm{~N}\). What was the initial speed \(t_{0}\) of the bullct?

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