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Chapter 8: Problem 32

A \(1050-\mathrm{kg}\) sports car is moving westbound at 15.0 \(\mathrm{m} / \mathrm{s}\) on a level road when it collides with a 6320 \(\mathrm{kg}\) truck driving east on the same road at 10.0 \(\mathrm{m} / \mathrm{s}\) . The two vehicles remain locked together after the collision. (a) What is the velocity (magnitude and direction) of the two vehicles just after the collision? (b) At what speed should the truck have been moving so that it and car are both stopped in the collision? (c) Find the change in kinetic energy of the system of two vehicles for the situations of part (a) and part (b). For which situation is the change in kinetic energy greater in magnitude?

Short Answer

Expert verified
(a) 6.44 m/s east. (b) 2.49 m/s east. (c) Greater change in part (b).

Step by step solution

01

Define the Problem and Known Variables

We have a collision problem where a 1050 kg sports car is moving west at 15 m/s and collides with a 6320 kg truck moving east at 10 m/s. After the collision, they stick together. We need to find the final velocity, a speed for a stopping collision, and kinetic energy changes.
02

Apply Conservation of Momentum

The total momentum before the collision should be the same as the total momentum after the collision. Calculate the momentum of each vehicle before the collision and set the total equal to the combined mass and velocity after the collision.Before collision:\[ p_{ ext{car}} = m_{ ext{car}} imes v_{ ext{car}} = 1050 imes (-15) = -15750 \, \mathrm{kg\ m/s} \]\[ p_{ ext{truck}} = m_{ ext{truck}} imes v_{ ext{truck}} = 6320 imes 10 = 63200 \, \mathrm{kg\ m/s} \]Net momentum before:\[ p_{ ext{total}} = -15750 + 63200 = 47450 \, \mathrm{kg\ m/s} \]
03

Solve for Final Velocity after Collision (Part a)

Use the conservation of momentum for the combined mass after the collision:\[ (m_{ ext{car}} + m_{ ext{truck}}) \cdot v_{ ext{final}} = p_{ ext{total}} \]\[ v_{ ext{final}} = \frac{47450}{1050 + 6320} = \frac{47450}{7370} \approx 6.44 \, \mathrm{m/s} \]Since the sum total momentum is positive, the velocity is eastbound.
04

Speed for Stop Collision (Part b)

Set the final velocity to zero, meaning total initial momentum should be zero:\[ -15750 + 6320 \cdot v = 0 \]\[ v = \frac{15750}{6320} \approx 2.49 \, \mathrm{m/s} \]So, the truck should have been moving at 2.49 m/s east for both vehicles to stop.
05

Calculate Change in Kinetic Energy for Part (a)

Initial total kinetic energy:\[ KE_{ ext{initial}} = \frac{1}{2} \cdot 1050 \cdot 15^2 + \frac{1}{2} \cdot 6320 \cdot 10^2 \]\[ = 118125 + 316000 = 434125 \, \mathrm{J} \]Final kinetic energy:\[ KE_{ ext{final}} = \frac{1}{2} \cdot 7370 \cdot 6.44^2 \]\[ \approx 152591.8 \, \mathrm{J} \]Change in kinetic energy:\[ \Delta KE = KE_{ ext{final}} - KE_{ ext{initial}} = 152591.8 - 434125 \approx -281533.2 \, \mathrm{J} \]
06

Calculate Change in Kinetic Energy for Part (b)

The vehicles stop moving after collision.Final kinetic energy is\[ KE_{ ext{final}} = 0 \, \mathrm{J} \]Change in kinetic energy:\[ \Delta KE = 0 - 434125 = -434125 \, \mathrm{J} \]
07

Compare Magnitude of Energy Changes

The change in energy for part (b) is 434125 J, which is greater in magnitude compared to 281533.2 J from part (a). So, part (b) involves a greater magnitude of energy change.

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Key Concepts

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

Kinetic Energy
Kinetic energy (\[ KE \] ) is the energy an object has due to its motion. It depends on two main factors: mass and velocity. The formula to calculate it is \[ KE = \frac{1}{2} m v^2 \], where \(m\) is mass and \(v\) is velocity. In collision physics, we often compare the kinetic energy before and after an event to understand how energy is transformed or conserved.
  • Kinetic energy is always a positive quantity, as both mass and velocity squared are positive.
  • An object with greater velocity or mass will have more kinetic energy.
  • In elastic collisions, kinetic energy is conserved, while in inelastic collisions, like the one described, some kinetic energy is lost, typically to heat and sound.
Calculating kinetic energy changes helps us study energy transfer. For example, in the given exercise, we evaluate the energy before and after collision to understand the energy dissipation.
Collision Physics
When studying collisions, it's important to consider the types of collision and how they affect energy and momentum. Collisions can be categorized as elastic or inelastic.

In an **elastic collision**, objects collide and separate apart, conserving both momentum and kinetic energy. In contrast, **inelastic collisions**, like the one in our exercise, involve objects sticking together post-collision and only momentum is conserved, while kinetic energy is not.
  • Elastic collisions: Objects bounce off with no loss in total kinetic energy.
  • Inelastic collisions: Objects clump together and transform part of kinetic energy to heat, or deformation energy.
  • Perfectly inelastic collisions: Special type of inelastic collision where maximum kinetic energy is lost.
Understanding these collision types helps us predict outcomes and measure variables, such as post-collision velocity and energy changes.
Momentum Calculation
Momentum is a key principle in collision physics that expresses the mass and velocity of an object. It is calculated using the formula \[ p = m \times v \], where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.

The conservation of momentum principle states that the total momentum before a collision must equal the total momentum after, assuming no external forces act on the system.
  • Momentum is a vector quantity, which means direction is important when calculating it.
  • Conservation of momentum is crucial for calculating unknowns, like final velocities after collision.
  • Even if kinetic energy isn't conserved (inelastic collisions), momentum always is.
By calculating the total momentum before and after the collision in the exercise, we're able to find out important details like the final velocity of the vehicles, showcasing the power of momentum in predicting motion behavior.

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Most popular questions from this chapter

The expanding gases that leave the muzzle of a rifle also contribute to the recoil. A 30 -caliber bullet has mass 0.00720 \(\mathrm{kg}\) and a speed of 601 \(\mathrm{m} / \mathrm{s}\) relative to the muzzle when fired from a rifle that has mass 2.80 kg. The loosely held rifle recoils at a speed of 1.85 \(\mathrm{m} / \mathrm{s}\) relative to the earth. Find the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle.

You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock so it has velocity 12.0 \(\mathrm{m} / \mathrm{s}\) relative to the earth at an angle of \(35.0^{\circ}\) above the borizontal. If your mass is 70.0 \(\mathrm{kg}\) and the rock's mass is 15.0 kg, what is your speed after you throw the rock (see Discussion Question \(Q 8.7\) ?

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

(a) Show that the kinetic energy \(K\) and the momentum magnitude \(p\) of a particle with mass \(m\) are related by \(K=p^{2} / 2 m .\) (b) A \(0.040-\mathrm{kg}\) cardinal (Richmondena cardinalis) and a \(0.145-\mathrm{kg}\) baseball have the same kinetic energy. Which has the greater magnitude of momentum? What is the ratio of the cardinal's magnitude of momentum to the bascball's? A \(700-N\) man and a \(450-N\) woman have the same momentum. Who has the greater kinetic energy? What is the ratio of the man's kinetic energy to that of the woman?

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