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8 A car of mass \(m\) moving at a speed \(v_{1}\) collides and couples with the back of a truck of mass 2 m moving initially in the same direction as the car at a lower speed \(v_{2}\) - (a) What is the speed \(y_{f}\) of the two vehicles immediately after the collision? (b) What is the change in kinetic energy of the car-truck system in the collision?

Short Answer

Expert verified
The final speed \(v_{f}\) of the two vehicles immediately after collision is \(v_{f} = (m*v_{1} + 2m*v_{2}) / (m + 2m)\) and the change in kinetic energy of the car-truck system in the collision is \( ΔKE = KE_{f} - KE_{i}\) with \( KE_{f} = 1/2*(m+2m)*v_{f}^2 \) and \( KE_{i} = 1/2*m*v_{1}^2 + 1/2*2m*v_{2}^2 \)

Step by step solution

01

Apply conservation of momentum

We start the exercise by applying the conservation of momentum. According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. The equation for this will be \( m*v_{1} + 2m*v_{2} = (m + 2m)*v_{f} \). The only unknown in this equation is \( v_{f} \), which we can solve for.
02

Calculate the final velocity

We can now rearrange the equation to solve for \( v_{f} \). We will divide both sides of the equation by the total mass after collision which results in: \( v_{f} = (m*v_{1} + 2m*v_{2}) / (m + 2m) \)
03

Calculate the initial and final kinetic energy

We need to calculate the initial and final kinetic energy of the system to determine the change in kinetic energy. The kinetic energy before collision is:\( KE_{i} = 1/2*m*v_{1}^2 + 1/2*2m*v_{2}^2 \), and the kinetic energy after collision is:\( KE_{f} = 1/2*(m+2m)*v_{f}^2 \).
04

Find the change in kinetic energy

The change in kinetic energy (ΔKE) is the final kinetic energy subtract the initial kinetic energy: \( ΔKE = KE_{f} - KE_{i} \)

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Key Concepts

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

Understanding Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. Every moving object has kinetic energy, which can be calculated using the formula:\[KE = \frac{1}{2}mv^2\]Here, \(m\) represents mass and \(v\) represents velocity. This formula shows that kinetic energy increases with both the mass and the square of the velocity. The fact that velocity is squared in this equation means that speed changes have a significant impact on kinetic energy.

In collisions, kinetic energy might change before and after the event. It's worth noting that while the momentum is always conserved in collisions, kinetic energy might not be. For example, in inelastic collisions like in our problem, kinetic energy is not conserved while momentum is. This is why you'll often see the calculation of kinetic energy before and after a collision as a way to determine the amount of energy lost in the form of sound, heat, or deformation of the objects involved.
Types of Collisions
A collision occurs when two or more bodies exert forces on each other for a short period of time. There are different types of collisions based on how kinetic energy is conserved:
  • Elastic Collision: Both momentum and kinetic energy are conserved. The bodies bounce off each other without any loss of kinetic energy.
  • Inelastic Collision: Momentum is conserved, but kinetic energy is not. This type includes cases like our car hitting the back of the truck, where the bodies might stick together (coupling) or there's deformation involved.
  • Perfectly Inelastic Collision: This is the most extreme case of an inelastic collision, where the maximum possible kinetic energy is lost. The colliding objects move together as one mass after the collision.
In the provided exercise, the car and truck collision is inelastic. This means that after they collide, some kinetic energy is transformed into other forms, such as heat or sound, and the two vehicles move together as a single object.
Conservation of Momentum
Momentum is a fundamental concept in physics and is defined as the product of an object's mass and velocity. The principle of conservation of momentum states that if no external forces are acting on a system, its total momentum remains constant. Mathematically, momentum \(p\) is expressed as:\[p = mv\]In our car and truck problem, momentum before the collision is calculated by adding the momentum of each vehicle:\[(m \times v_1) + (2m \times v_2)\]After the collision, because the car and truck couple together, we consider them as a single mass with a combined velocity \(v_f\). The equation for momentum conservation is:\[(m \times v_1) + (2m \times v_2) = (m + 2m) \times v_f\]By applying this principle, we can solve for \(v_f\), the velocity of the combined mass after the collision. This step is crucial as it links momentum conservation to the observed changes post-collision, enabling us to predict the resulting motion of both vehicles.

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Most popular questions from this chapter

An archer shoors an arrow toward a \(300-g\) target that is sliding in her direction at a speed of \(2.50 \mathrm{~m} / \mathrm{s}\) on a smooth, slippery surface. The \(22.5-\mathrm{g}\) arrow is shot with a speed of \(35.0 \mathrm{~m} / \mathrm{s}\) and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target?

Two blocks of masses m1 and m2 approach each other on a horizontal table with the same constant speed, v0, as measured by a laboratory observer. The blocks undergo a perfectly elastic collision, and it is observed that m1 stops but m2 moves opposite its original motion with some constant speed, v. (a) Determine the ratio of the two masses, m1/m2. (b) What is the ratio of their speeds, v/v0?

S This is a symbolic version of Problem 33. A railroad car of mass \(M\) moving at a speed \(v_{1}\) collides and couples with two coupled railroad cars, each of the same mass \(M\) and moving in the same direction at a speed \(t_{2}\). (a) What is the speed \(v_{y}\) of the three coupled cars after the collision in terms of \(v_{1}\) and \(v_{2}\) ? (b) How much kinetic energy is lost in the collision? Answer in terms of \(M\), \(v_{1}\), and \(v_{2}\).

Calculate the magnitude of the linear momentum for the following cases: (a) a proton with mass equal to \(1.67 \times 10^{-27} \mathrm{~kg}\), moving with a speed of \(5.00 \times 10^{6}\) \(\mathrm{m} / \mathrm{s} ;\) (b) a \(15.0-\mathrm{g}\) bullet moving with a speed of \(300 \mathrm{~m} / \mathrm{s} ;\) (c) a \(75.0-\mathrm{kg}\) sprinter running with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\); (d) the Earth (mass \(=5.98 \times 10^{24} \mathrm{~kg}\) ) moving with an orbital speed equal to \(2.98 \times 10^{4} \mathrm{~m} / \mathrm{s}\).

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