/*! 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 49 A 2000 -kg car moving east at \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A 2000 -kg car moving east at \(10.0 \mathrm{~m} / \mathrm{s}\) collides with a 3000 -kg car moving north. The cars stick together and move as a unit after the collision, at an angle of \(40.0^{\circ}\) north of east and a speed of \(5.22 \mathrm{~m} / \mathrm{s}\). Find the speed of the \(3000-\mathrm{kg}\) car before the collision.

Short Answer

Expert verified
The speed of the 3000-kg car before the collision was \(7.81\,m/s\)

Step by step solution

01

Calculate the initial momentum

Calculate the initial momentum of the 2000-kg car moving east, which will be equal to its mass multiplied by its speed: \( momentum_{2000} = mass_{2000} * speed_{2000} = 2000\,kg * 10.0\, m/s = 20000\, kg·m/s \)
02

Calculate the final eastward and northward momentum

The final momentum of the combined cars in the eastward direction can be calculated from the resultant momentum and the angle it makes with the east direction. Similarly, the northward momentum can also be calculated. \( momentum_{east} = total\_mass * final\_speed * cos(40.0 degrees) = (2000\,kg + 3000\,kg) * 5.22\, m/s * cos(40.0) = 29016\,kg·m/s \) and \( momentum_{north} = total\_mass * final\_speed * sin(40.0 degrees) = (2000\,kg + 3000\,kg) * 5.22\, m/s * sin(40.0) = 23421\,kg·m/s \)
03

Apply the principle of conservation of momentum

Define the north direction as positive y direction and the east direction as positive x direction. The momentum before and after collision in each direction must be equal. Hence, in the x (east) direction, \( momentum_{2000} = momentum_{east} \) which implies that the initial velocity \( velocity_{3000x} = 0 \) as only 2000-kg car was moving in east direction. In the y (north) direction, the 3000-kg car was moving, \( velocity_{3000y} = momentum_{north} / mass_{3000} = 23421\, kg·m/s / 3000\,kg = 7.81\,m/s \)
04

Calculate the initial speed of 3000-kg car

Finally, we need to calculate the speed of the 3000-kg car before the collision. Since it was only moving in the y (north) direction, its speed is equal to its velocity in y direction. Thus, \( speed_{3000} = velocity_{3000y} = 7.81\,m/s \)

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics associated with moving objects. It's defined as the product of an object's mass and its velocity (speed with direction). Mathematically, this is expressed as
\[ momentum = mass \times velocity \]
Momentum is a vector quantity, which means it has both magnitude and direction. It plays a crucial role in understanding how objects interact in collisions. One of the key principles involving momentum is the conservation of momentum, which states that in the absence of external forces, the total momentum of a system remains constant before and after a collision.
In the exercise provided, we calculate the momentum of the 2000-kg car by multiplying its mass with its speed eastward. Understanding momentum helps us to predict the resultant velocity of objects after they collide, as shown when calculating the speed of the 3000-kg car before the collision.
Inelastic Collision
An inelastic collision is a type of collision where the colliding objects stick together and move as a single unit after the collision. Unlike elastic collisions, kinetic energy is not conserved in inelastic collisions; however, momentum is conserved.
This conservation of momentum is the key to solving many collision problems. During an inelastic collision, the velocities of the objects before the collision can be determined by using the principle of conservation of momentum, since the masses and the velocity of the combined object afterward are usually known. It is essential to note that in these sorts of problems we often neglect external forces like friction or air resistance.
The exercise illustrates an inelastic collision where after impact, the cars stick together. By using the conservation of momentum in both the eastward and northward directions, we can deduce the original speed of the 3000-kg car, because the final velocity and direction of the combined mass after the collision were known.
Vector Components
When dealing with physical quantities that have a direction, such as momentum or velocity, vector components become very useful. A vector can be broken down into its component parts, typically along the x (horizontal) and y (vertical) axes. This is significant when calculating the result of vector-based operations like adding velocities or momenta in multiple directions.

Breaking Down Vectors into Components

To find the components of a vector, we use trigonometric functions based on the angle the vector makes with the axes. The cosine of the angle gives us the x-component, and the sine gives us the y-component. In the context of our exercise, this was critical for determining the momentum in the eastward (x) and northward (y) directions after the collision. By multiplying the combined mass and final speed with the cosine and sine of the given angle, respectively, we can find these momentum components:
\[ momentum_{component} = mass \times speed \times trigonometric\function(angle) \]
The concept of vector components allows us to apply the conservation of momentum in multi-dimensional motion to find unknown quantities, such as the initial speed of the 3000-kg car in the northward direction before the collision.

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Most popular questions from this chapter

Two objects of masses \(m\) and \(3 m\) are moving toward each other along the \(x\) -axis with the same initial speed \(v_{0}\). The object with mass \(m\) is traveling to the left, and the object with mass \(3 m\) is traveling to the right. They undergo an elastic glancing collision such that \(m\) is moving downward after the collision at right angles from its initial direction. (a) Find the final speeds of the two objects. (b) What is the angle \(\theta\) at which the object with mass \(3 m\) is scattered?

IA 0.280-kg volleyball approaches a player horizontally with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\). The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of \(22.0 \mathrm{~m} / \mathrm{s}\). (a) What impulse is delivered to the ball by the player? (b) If the player's fist is in contact with the ball for \(0.0600 \mathrm{~s}\), find the magnitude of the average force exerted on the player's fist.

A \(7.0\) -g bullet is fired into a \(1.5\) -kg ballistic pendulum. The bullet emerges from the block with a speed of \(200 \mathrm{~m} / \mathrm{s}\), and the block rises to a maximum height of \(12 \mathrm{~cm}\). Find the initial speed of the bullet.

A A \(45.0\) -kg girl is standing on a 150 -kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of \(1.50 \mathrm{~m} / \mathrm{s}\) to the right relative to the plank. (a) What is her velocity relative to the surface of the ice? (b) What is the velocity of the plank relative to the surface of the ice?

ecp Show that the kinetic energy of a particle of mass \(m\) is related to the magnitude of the momentum \(p\) of that particle by \(K E=p^{2} / 2 m .\) Note: This expression is invalid for particles traveling at speeds near that of light.

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