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Chapter 14: Problem 8

Two identical cars \(A\) and \(B\) are at rest on a loading dock with brakes released. Car \(C,\) of a slightly different style but of the same weight, has been pushed by dockworkers and hits car \(B\) with a velocity of \(1.5 \mathrm{m} / \mathrm{s}\). Knowing that the coefficient of restitution is 0.8 between \(B\) and \(C\) and 0.5 between \(A\) and \(B\), determine the velocity of each car after all collisions have taken place.

Short Answer

Expert verified
Cars A and B are at 0.6 m/s, car C is at 0.7 m/s after collisions.

Step by step solution

01

Understanding the Problem

We have three cars: Car \( A \), \( B \), and \( C \). Cars \( A \) and \( B \) are initially at rest, while car \( C \) is moving with a velocity of \( 1.5 \, \text{m/s} \). Car \( C \) collides with car \( B \), and then car \( B \) collides with car \( A \). The coefficient of restitution between \( B \) and \( C \) is \( 0.8 \), and between \( A \) and \( B \) is \( 0.5 \). We need to find the final velocities of each car after all collisions.
02

Apply Conservation of Momentum for Collision (B and C)

For the collision between cars \( B \) and \( C \), initially, only car \( C \) is moving. By conservation of momentum, \( m_C v_{C1} = m_C v_{C2} + m_B v_{B2} \), where \( v_{C1} = 1.5 \, \text{m/s}\) and both masses are equal and can be canceled out. Solve for velocities after collision.
03

Use Coefficient of Restitution for Collision (B and C)

The coefficient of restitution formula is \( e = \frac{v_{B2} - v_{C2}}{v_{C1} - v_{B1}} \). Plug in \( e = 0.8 \), \( v_{C1} = 1.5 \, \text{m/s} \), and \( v_{B1} = 0 \, \text{m/s} \) to solve for \( v_{B2} \) and \( v_{C2} \).
04

Solve System of Equations from Steps 2 and 3

Using the equations from Steps 2 and 3, solve for \( v_{B2} \) and \( v_{C2} \):\[ v_{C2} = 0.7 \, \text{m/s} \]\[ v_{B2} = 1.2 \, \text{m/s} \]
05

Apply Conservation of Momentum for Collision (A and B)

Now, car \( B \) at \( 1.2 \, \text{m/s} \) collides with car \( A \) at rest. By conservation of momentum: \( m_B v_{B2} = m_A v_{A3} + m_B v_{B3} \). Solve this equation for the velocities \( v_{A3} \) and \( v_{B3} \).
06

Use Coefficient of Restitution for Collision (A and B)

The coefficient of restitution for cars \( A \) and \( B \) is \( 0.5 \). Using \( e = \frac{v_{B3} - v_{A3}}{v_{B2} - v_{A2}} \), and knowing \( v_{A2} = 0 \, \text{m/s} \), solve for \( v_{A3} \) and \( v_{B3} \).
07

Solve System of Equations from Steps 5 and 6

Using the two equations derived from the conservation of momentum and the coefficient of restitution, solve for \( v_{A3} \) and \( v_{B3} \):\[ v_{A3} = 0.6 \, \text{m/s} \]\[ v_{B3} = 0.6 \, \text{m/s} \]
08

Final Velocities

After both collisions, the final velocities of the cars are: \( v_A = 0.6 \, \text{m/s} \), \( v_B = 0.6 \, \text{m/s} \), and \( v_C = 0.7 \, \text{m/s} \).

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Key Concepts

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

Conservation of Momentum
The principle of conservation of momentum is crucial in understanding collisions.
Momentum is the product of an object's mass and velocity. During a collision, the total momentum before impact is equal to the total momentum after impact, provided no external forces interfere. This is expressed mathematically as:
  • \[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]
In our problem with cars, car C is initially moving with a velocity while A and B are at rest. As car C strikes car B, this law helps us calculate momentum redistribution.
Although the cars have the same mass, understanding how momentum shifts gives insight into post-collision velocities. Hence, for every collision, the rearrangement of initial momentum dictates the subsequent motions of the cars.
Collision Physics
Collisions can be complex, but the core principles remain consistent.
When two objects collide, there's an exchange of velocity and energy. Here, the coefficient of restitution (\( e \)) comes into play. This value represents the bounce-back ability between two colliding bodies. In simpler terms, it defines how 'bouncy' a collision is.
  • For collision between B and C: \( e = 0.8 \), indicating most of the kinetic energy is retained.
  • For A and B: \( e = 0.5 \), showing more energy is lost.
The equation for restitution relates velocities before and after collision:
  • \[ e = \frac{v_{after} - (-v_{before})}{v_{before} - 0} \]
For our scenario, these coefficients guide us in calculating post-collision velocities, reflecting energy conservation efficiency.
Kinematics
Kinematics deals with motion aspects without considering the causes.
In our context, it involves figuring out the velocities of the cars after each collision. Using the conservation of momentum and coefficient of restitution equations, we solve the final velocities mathematically.
Initially, only car C has motion, moving at \( 1.5 \; \text{m/s} \). When it strikes car B, kinematic principles help us compute resulting speeds.
  • For B after B-C collision: \( v_{B2} = 1.2 \; \text{m/s} \).
  • For C after B-C collision: \( v_{C2} = 0.7 \; \text{m/s} \).
As car B moves to hit car A, a new kinematic calculation determines their post-collision velocities, considering a reduced restitution coefficient, resulting in:
  • For A after A-B collision: \( v_{A3} = 0.6 \; \text{m/s} \).
  • For B after A-B collision: \( v_{B3} = 0.6 \; \text{m/s} \).
By analyzing velocities, kinematics gives a picture of the motion evolution due to collisions.

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Most popular questions from this chapter

Two identical 1350 -kg automobiles \(A\) and \(B\) are at rest with their brakes released when \(B\) is struck by a \(5400-\mathrm{kg}\) truck \(C\) that is moving to the left at \(8 \mathrm{km} / \mathrm{h}\). A second collision then occurs when \(B\) strikes \(A\). Assuming the first collision is perfectly plastic and the second collision is perfectly elastic, determine the velocities of the three vehicles just after the second collision.

An expert archer demonstrates his ability by hitting tennis balls thrown by an assistant. A 2 -oz tennis ball has a velocity of \((32 \mathrm{ft} / \mathrm{s}) \mathrm{i}-(7 \mathrm{ft} / \mathrm{s}) \mathrm{j}\) and is \(33 \mathrm{ft}\) above the ground when it is hit by \(\mathrm{a} 1.2-\mathrm{oz}\) arrow traveling with a velocity of \((165 \mathrm{ft} / \mathrm{s}) \mathrm{j}+(230 \mathrm{ft} / \mathrm{s}) \mathrm{k}\) where \(\mathrm{j}\) is directed upwards. Determine the position \(P\) where the ball and arrow will hit the ground, relative to point \(O\) located directly under the point of impact.

An 18 -kg cannonball and a 12 -kg cannonball are chained together and fired horizontally with a velocity of \(165 \mathrm{m} / \mathrm{s}\) from the top of \(\mathrm{a} 15-\mathrm{m}\) wall. The chain breaks during the flight of the cannon-balls and the \(12-\mathrm{kg}\) cannonball strikes the ground at \(t=1.5 \mathrm{s}\), at a distance of \(240 \mathrm{m}\) from the foot of the wall, and \(7 \mathrm{m}\) to the right of the line of fire. Determine the position of the other cannonball at that instant. Neglect the resistance of the air.

Two small disks \(A\) and \(B\), of mass \(2 \mathrm{kg}\) and \(1 \mathrm{kg}\), respectively, may slide on a horizontal and frictionless surface. They are connected by a cord of negligible mass and spin about their mass center \(G . \mathrm{At} t=0\), \(G\) is moving with the velocity \(\overline{\mathrm{v}}_{0}\) and its coordinates are \(\bar{x}_{0}=0, \bar{y}_{0}=1.89 \mathrm{m} .\) Shortly thereafter, the cord breaks and disk \(A\) is observed to move with a velocity \(\mathbf{v}_{A}=(5 \mathrm{m} / \mathrm{s}) \mathrm{j}\) in a straight line and at a distance \(a=2.56 \mathrm{m}\) from the \(y\) axis, while \(B\) moves with a velocity \(\mathrm{v}_{B}=(7.2 \mathrm{m} / \mathrm{s}) \mathrm{i}-(4.6 \mathrm{m} / \mathrm{s}) \mathrm{j}\) along a path intersecting the \(x\) axis at \(\mathrm{a}\) distance \(b=7.48 \mathrm{m}\) from the origin \(0 .\) Determine (a) the initial velocity \(\overline{\mathbf{v}}_{0}\) of the mass center \(G\) of the two disks, \((b)\) the length of the cord initially connecting the two disks, (c) the rate in rad/s at which the disks were spinning about \(G .\)

In a Pelton-wheel turbine, a stream of water is deflected by a series of blades so that the rate at which water is deflected by the blades is equal to the rate at which water issues from the nozzle \(\Delta m / \Delta t=\) \(A \rho v_{d}\). Using the same notation as in Sample Prob. \(14.8,(a)\) determine the velocity \(V\) of the blades for which maximum power is developed, \((b)\) derive an expression for the maximum power, \((c)\) derive an expression for the mechanical efficiency.
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