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Chapter 3: Problem 243

Car \(B\) is initially stationary and is struck by car \(A\) which is moving with speed \(v\). The mass of car \(B\) is \(p m,\) where \(m\) is the mass of car \(A\) and \(p\) is a positive constant. If the coefficient of restitution is \(e=\) \(0.1,\) express the speeds \(v_{A}^{\prime}\) and \(v_{B}^{\prime}\) of the two cars at the end of the impact in terms of \(p\) and \(v .\) Evaluate your expressions for \(p=0.5\).

Short Answer

Expert verified
For \(p=0.5\), \(v'_A\approx 0.633v\) and \(v'_B\approx 0.7v\).

Step by step solution

01

Understand the Problem

We are given two cars, A and B. Car A is moving with initial speed \(v\) while car B is stationary. The mass of car B is given as \(pm\) where \(m\) is the mass of car A. The coefficient of restitution is \(e = 0.1\). We need to find the final velocities of both cars, \(v'_A\) and \(v'_B\), after the collision in terms of \(p\) and \(v\). Finally, evaluate these expressions for \(p = 0.5\).
02

Apply Conservation of Momentum

Since there are no external forces, the total linear momentum is conserved during the collision. The initial momentum is \(m \times v\) (momentum of car A) and the initial momentum of car B is zero. Thus, the total initial momentum is \(mv\). For the final state, the momentum is \(mv'_A + pmv'_B\). Hence, the equation for conservation of momentum is:\[mv = mv'_A + pmv'_B\]
03

Use the Coefficient of Restitution

The coefficient of restitution is a measure of the elasticity of the collision and is defined by the equation:\[e = \frac{v'_B - v'_A}{v - 0}\]Substituting \(e = 0.1\), we have:\[0.1 = \frac{v'_B - v'_A}{v}\]
04

Solve the Equations Simultaneously

We now have two equations:1. \(mv = mv'_A + pmv'_B\)2. \(0.1v = v'_B - v'_A\)Solving these simultaneously, start by substituting \(v'_B = v'_A + 0.1v\) from the second equation into the first equation:\[mv = mv'_A + pm(v'_A + 0.1v)\]Simplifying, we get:\[mv = v'_A(m + pm) + 0.1p mv\]\[v'_A = \frac{mv - 0.1pmv}{m(1 + p)}\]\[v'_A = \frac{v(1 - 0.1p)}{1 + p}\]Substitute \(v'_A\) back to find \(v'_B\):\[v'_B = v'_A + 0.1v = \frac{v(1 - 0.1p)}{1 + p} + 0.1v\]
05

Simplify and Evaluate for p=0.5

Now, simplify \(v'_B\):\[v'_B = \frac{v(1 - 0.1p)}{1 + p} + 0.1v = \frac{v(1 - 0.1p) + 0.1v(1 + p)}{1 + p}\]\[v'_B = \frac{v(1 - 0.1p + 0.1 + 0.1p)}{1 + p} = \frac{v(1 + 0.1p)}{1 + p}\]For \(p = 0.5\), evaluate both velocities:\[v'_A = \frac{v(1 - 0.1 \times 0.5)}{1 + 0.5} = \frac{v(1 - 0.05)}{1.5} = \frac{0.95v}{1.5}\approx 0.633v\]\[v'_B = \frac{v(1 + 0.1 \times 0.5)}{1 + 0.5} = \frac{v(1 + 0.05)}{1.5} = \frac{1.05v}{1.5}\approx 0.7v\]

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Key Concepts

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

Coefficient of Restitution
The coefficient of restitution is a crucial concept in understanding collision dynamics. It tells us how bouncy or elastic a collision is. Essentially, it's a measurement of the speed loss during collision. This coefficient is defined as the ratio of the relative velocity of separation to the relative velocity of approach.
For example, in the exercise, it is given as:
  • Coefficient of restitution, \( e = 0.1 \)
  • Relative velocity of separation = \( v'_B - v'_A \)
  • Relative velocity of approach = \( v \)
Thus, the equation becomes:\[ e = \frac{v'_B - v'_A}{v} \]Here, if \( e = 1 \), the collision is perfectly elastic, meaning no kinetic energy is lost during the impact. Conversely, if \( e = 0 \), the collision is perfectly inelastic, showing that all kinetic energy during the collision is lost. For our exercise, the small \( e \) value (0.1) suggests the collision leans heavily towards being inelastic. This means there is a significant loss of speed after the collision.
Conservation of Momentum
According to the law of conservation of momentum, the total momentum of a system remains constant, provided no external forces act on it. Momentum is the product of mass and velocity, and for our exercise, we have two cars colliding.
The conservation of momentum equation is set as follows:
  • Initial momentum: \( mv \) for car A (since car B is stationary)
  • Final momentum: \( mv'_A + pmv'_B \)
To conserve momentum, the following equation should hold:\[ mv = mv'_A + pmv'_B \]This equation allows us to solve for one of the unknowns, given the mass and velocity of each car. Since we have no external forces affecting the motion of the cars, their combined momentum remains consistent before and after the collision. This fundamental principle helps in solving for the velocities after impact.
Collision Dynamics
Collision dynamics is the study of physical changes during a collision between bodies, focusing on both motion and forces. In our scenario, we deal with two vehicles colliding, requiring us to pay attention to their masses, velocities, and how energy moves between them.
Understanding collision dynamics encompasses:
  • Mass distribution: Car A has mass \( m \), and car B has mass \( pm \), affecting how the impact distributes between them.
  • Energy transfer: The energy transfer depends on the coefficient of restitution and the masses.
  • Direction and speed post-collision: Calculating \( v'_A \) and \( v'_B \) based on given values.
Through the combined use of the coefficient of restitution and conservation of momentum, we determine the velocities after impact. Calculating the dynamics of the collision for different masses, such as when \( p = 0.5 \), allows us to see real-world application by obtaining the speeds as fractions of the initial velocity: \( v'_A \approx 0.633v \) and \( v'_B \approx 0.7v \). These calculations illustrate the physical dynamics at play within differing conditions and ensure a comprehensive understanding of vehicle collisions.

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Most popular questions from this chapter

The aircraft carrier is moving at a constant speed and launches a jet plane with a mass of \(3 \mathrm{Mg}\) in a distance of \(75 \mathrm{m}\) along the deck by means of a steam-driven catapult. If the plane leaves the deck with a velocity of \(240 \mathrm{km} / \mathrm{h}\) relative to the carrier and if the jet thrust is constant at \(22 \mathrm{kN}\) during takeoff, compute the constant force \(P\) exerted by the catapult on the airplane during the \(75-\mathrm{m}\) travel of the launch carriage.

The bungee jumper, an 80 -kg man, falls from the bridge at \(A\) with the bungee cord secured to his ankles. He falls \(20 \mathrm{m}\) before the 17 -m length of elastic bungee cord begins to stretch. The \(3 \mathrm{m}\) of rope above the elastic cord has no appreciable stretch. The man is observed to drop a total of \(44 \mathrm{m}\) before being projected upward. Neglect any energy loss and calculate \((a)\) the stiffness \(k\) of the bungee cord (increase in tension per meter of elongation) (b) the maximum velocity \(v_{\max }\) of the man during his fall, and \((c)\) his maximum acceleration \(a_{\max } .\) Treat the man as a particle located at the end of the bungee cord.

The third and fourth stages of a rocket are coasting in space with a velocity of \(18000 \mathrm{km} / \mathrm{h}\) when a small explosive charge between the stages separates them. Immediately after separation the fourth stage has increased its velocity to \(v_{4}=18060 \mathrm{km} / \mathrm{h} .\) What is the corresponding velocity \(v_{3}\) of the third stage? At separation the third and fourth stages have masses of 400 and \(200 \mathrm{kg},\) respectively.

The simple 2 -kg pendulum is released from rest in the horizontal position. As it reaches the bottom position, the cord wraps around the smooth fixed pin at \(B\) and continues in the smaller arc in the vertical plane. Calculate the magnitude of the force \(R\) supported by the pin at \(B\) when the pendulum passes the position \(\theta=30^{\circ}\).

Car \(A\) weighing 3200 lb and traveling north at \(20 \mathrm{mi} / \mathrm{hr}\) collides with car \(B\) weighing \(3600 \mathrm{lb}\) and traveling at \(30 \mathrm{mi} / \mathrm{hr}\) as shown. If the two cars become entangled and move together as a unit after the crash, compute the magnitude \(v\) of their common velocity immediately after the impact and the angle \(\theta\) made by the velocity vector with the north direction.
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