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# Physics and Collisions Help (page 2)

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## Two More Noteworthy Items

Two things ought to be mentioned before we continue on. First, until now, units have been included throughout calculations for illustrative purposes. Units can be multiplied and divided, just as can numbers. For example, 0.850 kg · m/s divided by 4.10 kg causes kilograms to cancel out in the final result, yielding meters per second (m/s). It’s a good idea to keep the units in calculations, at least until you get comfortable with them, so that you can be sure that the units in the final result make sense. If we had come up with, say, kilogram-meters (kg · m) in the final result for Problem 1, we would know that something was wrong because kilogram-meters are not units of speed or velocity magnitude.

The second thing you should know is that it’s perfectly all right to multiply and divide vector quantities, such as velocity or momentum, by scalar quantities, such as mass. This always yields another vector. For example, in the solution of Problem 8-2, we divided momentum (a vector) by mass (a scalar). However, we cannot add a vector to a scalar so easily or subtract a scalar from a vector. Nor can we multiply two vectors and expect to get a meaningful answer unless we define whether we are to use the dot product or the cross product. You should be familiar with this from your high school mathematics courses. If not, go to Part Zero of this book and review the material on vectors.

## Collisions Practice Problems

#### Problem 1

Suppose that you have two toy electric trains set up on a long, straight track running east and west. Train A has a mass 1.60 kg and travels east at 0.250 m/s. Train B has a mass of 2.50 kg and travels west at 0.500 m/s. The trains have stick pads on the fronts of their engines so that if they crash, they will not bounce off each other. The trains are set up so that they will crash. Suppose that the friction in the wheel bearings is zero, and suppose that the instant the trains hit each other, you shut off the power to the engines. How fast and in what direction will the composite train be moving after the crash? Assume that neither train derails.

#### Solution 1

First, calculate the momentum of each train. Call the masses of the trains m a and m b , respectively. Let us assign the directions east as positive and west as negative. (We can do this because they’re exactly opposite each other along a straight line.) Let the velocity vector of train A be represented as v a and the velocity vector of train B be represented as v b . Then m a = 1.60 kg, m b = 2.50 kg, v a = +0.250 m/s, and v b = −0.500 m/s. Their momentums, respectively, are

p a = m a v a = (1.60 kg)( + 0.250 m/s) = + 0.400 kg · m/s

p b = m b v b = (2.50 kg)(−0.500 m/s) = −1.25 kg · m/s

The sum total of their momentums is therefore

p = p a + p b = + 0.400 kg · m/s + (−1.25 kg · m/s)

= −0.850 kg · m/s

The mass m of the composite is simply the sum of the masses of trains A and B , which remain the same throughout this violent process:

m = m a + m b = 1.60 kg + 2.50 kg = 4.10 kg

Let the final velocity, for which we are trying to solve, be denoted v . We know that momentum is conserved in this collision, as it is in all ideal collisions. Therefore, the final velocity must be equal to the final momentum p divided by the final mass m :

v = p / m = (−0.850 kg · m/s)/(4.10 kg)

≈ −0.207 m/s

This means the composite “train,” after the crash, will move west at 0.207 m/s.

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