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# Physics and Collisions Help (page 3)

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#### Problem 2

Suppose that you have two toy electric trains set up as in Problem 1. Train A has a mass 2.00 kg and travels east at 0.250 m/s. Train B has a mass of 1.00 kg and travels west at 0.500 m/s. How fast and in what direction will the composite train be moving after the crash? Assume that neither train derails.

#### Solution 2

Call the masses of the trains m a and m b , respectively. Assign the directions east as positive and west as negative. Let the velocity vectors be represented as v a and v b . Then m a = 2.00 kg, m b = 1.00 kg, v a = +0.250 m/s, and v b = −0.500 m/s. Their momentums, respectively, are

P a = m a v a = (2.00 kg)(+ 0.250 m/s) = + 0.500 kg · m/s

p b = m b v b = (1.00 kg)(−0.500 m/s) = −0.500 kg · m/s

The sum total of their momentums is therefore

p = p a + p b = + 0.500 kg · m/s + (−0.500 kg · m/s)

= 0 kg · m/s

The mass m of the composite is simply the sum of the masses of trains A and B , neither of which change:

m = m a + m b = 2.00 kg + 1.00 kg = 3.00 kg

The final velocity v is equal to the final momentum p divided by the final mass m :

v = p / m = (0 kg · m/s)/(3.00 kg)

= 0 m/s

This means that the composite “train,” after the crash, is at rest. This might at first seem impossible. If momentum is conserved, how can it be zero after the crash? Where does it all go? The answer to this question is that the total momentum of this system is zero before the crash as well as after. Remember that momentum is a vector quantity. Look at the preceding equations again:

p a = m a v a = (2.00 kg)(+0.250 m/s) = +0.500 kg · m/s

p b = m b v b = (1.00 kg)(−0.500 m/s) = −0.500 kg · m/s

The momentums of the trains have equal magnitude but opposite direction. Thus their vector sum is zero before the crash.

Practice problems of this concept can be found at: Momentum, Work, Energy, And Power Practice Test

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