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Chapter 6: Problem 33

A railroad car of mass \(2.00 \times 10^{4} \mathrm{~kg}\) moving at \(3.00 \mathrm{~m} / \mathrm{s}\) collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at \(1.20 \mathrm{~m} / \mathrm{s}\). (a) What is the speed of the three coupled cars after the collision? (b) How much kinetic energy is lost in the collision?

Short Answer

Expert verified
The final speed of the three coupled cars after the collision (a) and how much kinetic energy is lost in the collision (b) can be calculated using the steps described above. The answers obtained will be numerical values dependent on the calculations made in the steps.

Step by step solution

01

Conservation of Momentum

To solve (a), we'll use the principle of conservation of momentum, which states that the total momentum before the collision should be equal to the total momentum after the collision. Mathematically, it can be represented as \(m_1v_1 + m_2v_2 = (m_1 + m_2 + m_3)v\), where \(m_1, m_2, m_3\) are the masses of the cars and \(v_1, v_2, v\) are the velocities before and after the collision respectively.
02

Calculating Final Velocity

Substituting the known values into the equation we have \(2.00 \times 10^{4} \mathrm{~kg} \times 3.00 \mathrm{~m/s} + 2 \times (2.00 \times 10^{4} \mathrm{~kg} \times 1.20 \mathrm{~m/s}) = (2.00 \times 10^{4} \mathrm{~kg} + 2 \times 2.00 \times 10^{4} \mathrm{~kg}) \times v\). Solving this will give us the final velocity \(v\) of the coupled cars.
03

Initial and Final Kinetic Energy

To solve (b), we need to first calculate the initial kinetic energy, given by \(K.E._1 = \frac{1}{2} m v^2\). The final kinetic energy will be given by \(K.E._2 = \frac{1}{2} M v_f^2\), where \(M\) is the total mass after collision and \(v_f\) is the final velocity obtained in step 2.
04

Calculating Energy Lost

Once we obtain the initial and the final kinetic energy, the energy lost in the form of heat, sound etc. during the collision can be obtained by taking the difference: Energy Lost = Initial Kinetic Energy - Final Kinetic Energy.

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Most popular questions from this chapter

Two blocks collide on a frictionless surface. After the collision, the blocks stick together. Block A has a mass M and is initially moving to the right at speed v. Block B has a mass 2M and is initially at rest. System C is composed of both blocks. (a) Draw a force diagram for each block at an instant during the collision. (b) Rank the magnitudes of the horizontal forces in your diagram. Explain your reasoning. (c) Calculate the change in momentum of block A, block B, and system C. (d) Is kinetic energy conserved in this collision? Explain your answer. (This problem is courtesy of Edward F. Redish. For more such problems, visit http://www.physics.umd.edu/perg.)

Q|C Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of \(12 \mathrm{~m} / \mathrm{s}\), and the mass of rain per second striking the roof is \(0.035 \mathrm{~kg} / \mathrm{s}\). (a) Assuming the drops come to rest after striking the roof, find the average force exerted by the rain on the roof. (b) If hailstones having the same mass as the raindrops fall on the roof at the same rate and with the same speed, how would the average force on the roof compare to that found in part (a)?

A high-speed photograph of a club hitting a golf ball is shown in Figure \(6.3\). The club was in contact with a ball, initially at rest, for about \(0.0020 \mathrm{~s}\). If the ball has a mass of \(55 \mathrm{~g}\) and leaves the head of the club with a speed of \(2.0 \times 10^{2} \mathrm{ft} / \mathrm{s}\), find the average force exerted on the ball by the club.

A ball of mass \(0.150 \mathrm{~kg}\) is dropped from rest from a height of \(1.25 \mathrm{~m}\). It rebounds from the floor to reach a height of \(0.960 \mathrm{~m}\). What impulse was given to the ball by the floor?

This is a symbolic version of Problem 23. A girl of mass \(m_{G}\) is standing on a plank of mass \(m_{p}\). Both are originally at rest on a frozen lake that constitutes a frictionless, flat surface. The girl begins to walk along the plank at a constant velocity \(x_{C P}\) to the right relative to the plank. (The subscript \(G P\) denotes the girl relative to plank.) (a) What is the velocity \(v_{P}\) of the plank relative to the surface of the ice? (b) What is the girl's velocity \(v_{C I}\) relative to the ice surface?
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