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Chapter 7: Problem 32

A car (mass \(=1100 \mathrm{kg}\) is traveling at 32 \(\mathrm{m} / \mathrm{s}\) when it collides head- on with a sport utility vehicle (mass \(=2500 \mathrm{kg}\) traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling?

Short Answer

Expert verified
The SUV was traveling at 14.08 m/s before the collision.

Step by step solution

01

Understand Conservation of Momentum

In any collision, the total momentum before the collision is equal to the total momentum after the collision, provided no external forces act on the system. This is known as the conservation of momentum.
02

Identify the Known Quantities

We are given the masses and velocities of the car and the sport utility vehicle (SUV). For the car: \( m_1 = 1100 \; \mathrm{kg}, v_1 = 32 \; \mathrm{m/s} \). The SUV's mass is \( m_2 = 2500 \; \mathrm{kg} \), but its velocity \( v_2 \) is unknown. Both vehicles come to a halt post-collision, so the final velocities are 0.
03

Set Up the Conservation of Momentum Equation

The momentum before the collision (\( m_1 \cdot v_1 + m_2 \cdot v_2 \)) is equal to the momentum after the collision (which is zero because both vehicles come to rest). Therefore, the equation is:\[1100 \cdot 32 + 2500 \cdot (-v_2) = 0\]
04

Solve for the Sport Utility Vehicle's Velocity

Rearrange the equation to solve for \( v_2 \):\[1100 \cdot 32 = 2500 \cdot v_2\]Divide both sides by 2500 to isolate \( v_2 \):\[v_2 = \frac{1100 \cdot 32}{2500}\]Calculate \( v_2 \) to find the SUV's velocity.
05

Calculate the Result

Perform the calculation:\[v_2 = \frac{1100 \times 32}{2500} = \frac{35200}{2500} = 14.08 \; \mathrm{m/s}\]Therefore, the initial velocity of the SUV was approximately 14.08 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 that describes the quantity of motion an object possesses. It is calculated as the product of an object's mass and velocity:
  • Momentum formula: \ \( \text{momentum} = m \times v \ \, \)
  • where \( m \) is the mass of the object, and \( v \) is the velocity.
Momentum is a vector quantity, meaning it has both magnitude and direction. This is important in problems involving collisions because the direction of momentum must be considered.

In the context of a collision, the total momentum of the involved objects before the collision must equal their total momentum after, assuming no external forces are at work. This principle is called conservation of momentum.
Collision
A collision occurs when two or more bodies exert forces on each other for a relatively short time. Collisions can be categorized into various types such as elastic and inelastic, and each type follows different principles:
  • Elastic collisions conserve both momentum and kinetic energy.
  • Inelastic collisions conserve momentum but not kinetic energy.
In the exercise at hand, we are dealing with an inelastic collision where both vehicles come to a complete stop. This means that the kinetic energy is not conserved, but the momentum of the system is.

In such situations, applying the conservation of momentum helps to analyze and solve for unknowns in the system, such as the initial velocity of one of the vehicles involved, by using the known quantities from the exercise.
Velocity
Velocity is a key factor in determining momentum because it describes how fast an object moves and in what direction. Unlike speed, which is only about how fast something is going, velocity also tells us the direction of travel:
  • Velocity includes both magnitude and direction.
  • A positive or negative sign usually indicates direction, positive forward and negative for the opposite direction.
In collision scenarios, direction is crucial because colliding objects often move towards each other, meaning their velocities are in opposite directions.

In the exercise, the known velocity of the car is 32 m/s, and the SUV鈥檚 velocity is unknown. Applying these values in the conservation of momentum equation allows us to compute the unknown velocity of the SUV.
Mass
Mass is another fundamental quantity crucial to momentum calculations. It represents the amount of matter in an object and is a scalar quantity, which means it only has magnitude and no direction:
  • Generally measured in kilograms (kg).
  • Larger mass typically results in larger momentum for a given velocity.
In the problem exercise, two different masses are involved: the car with 1100 kg and the SUV with 2500 kg. The disparity in these masses significantly affects the system's momentum.

Knowing the mass of each object allows us to set up the momentum equation properly and solve for unknowns, using the conservation of momentum principle to understand how the different amounts of mass influence the final velocities of colliding objects.

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Most popular questions from this chapter

A student (m 63 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.040 s. The average force exerted on him by the ground is 18 000 N, where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.

A model rocket is constructed with a motor that can provide a total impulse of 29.0 N s. The mass of the rocket is 0.175 kg. What is the speed that this rocket achieves when launched from rest? Neglect the effects of gravity and air resistance.

An electron collides elastically with a stationary hydrogen atom. The mass of the hydrogen atom is 1837 times that of the electron. Assume that all motion, before and after the collision, occurs along the same straight line. What is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision?

mmh For tests using a ballistocardiograph, a patient lies on a horizontal platform that is supported on jets of air. Because of the air jets, the friction impeding the horizontal motion of the plafform is negligible. Each time the heart beats, blood is pushed out of the heart in a direction that is nearly parallel to the platform. Since momentum must be conserved, the body and the platform recoil, and this recoil can be detected to provide information about the heart. For each beat, suppose that 0.050 \(\mathrm{kg}\) of blood is pushed out of the heart with a velocity of \(+0.25 \mathrm{m} / \mathrm{s}\) and that the mass of the pattient and platform is 85 \(\mathrm{kg}\) . Assuming that the patient does not slip with respect to the platform, and that the patient and platform start from rest, determine the recoil velocity.

A 60.0 -kg person, running horizontally with a velocity of \(+3.80 \mathrm{m} / \mathrm{s}\) jumps onto a \(12.0-\mathrm{kg}\) sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast 30.0 \(\mathrm{m}\) on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?
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