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Chapter 9: Problem 43

An 11,000 -kg freight car rests against a spring bumper at the end of a railroad track. The spring has constant \(k=0.32 \mathrm{MN} / \mathrm{m}\). The car is hit by a second car of \(9400-\mathrm{kg}\) mass moving at \(8.5 \mathrm{m} / \mathrm{s}\), and the two couple together. Find (a) the maximum compression of the spring and (b) the speed of the two cars when they rebound together from the spring.

Short Answer

Expert verified
The maximum compression of the spring is \(4.95 m\) and the speed of the two cars when they rebound from the spring is \(4.97 m/s\).

Step by step solution

01

Calculate the initial momentum

Calculate the initial momentum of the second car as it is the only one moving. The formula for momentum is \(P = mv\), where \(m\) is the mass and \(v\) is the velocity. Substituting the given values: \(P = (9400 kg) * (8.5 m/s) = 79900 kg*m/s.\)
02

Calculate the velocity after the collision

The total momentum before and after the collision will be equal. The formula for momentum is \(P = mv\), and the two cars couple together to form a single mass. Substituting the given values: \(79900 kg*m/s = (11000 kg + 9400 kg) * v\), we get \(v = 4.97 m/s\). This is the velocity after the collision and the velocity with which the coupled cars hit the spring.
03

Calculate the maximum compression of the spring

The maximum compression of the spring occurs when the coupled cars come to rest momentarily, at which point all their kinetic energy is transferred to potential energy in the spring. The formula for potential energy stored in a spring is \(PE = 1/2 * k * x^2\), and the kinetic energy is \(KE = 1/2 * mv^2\). Setting these equal to each other to calculate \(x\): \(1/2 * (20000 kg) * (4.97 m/s)^2 = 1/2 * ((0.32 MN/m) * 10^6 N/MN) * x^2\), yields \(x = 4.95 m\).
04

Calculate the speed of the cars when they rebound

As the spring expands back to its natural length, all the potential energy stored in it is transferred back into kinetic energy and the cars rebound. The speed of the cars when they rebound is the same speed they had when they first compressed the spring, since the energy stored is the same. Therefore the rebound speed is \(4.97 m/s\).

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