Introduction to Collisions
When two objects strike each other because they are in relative motion and their paths cross at exactly the right time, a collision is said to occur.
Conservation Of Momentum
According to the law of conservation of momentum , the total momentum contained in two objects is the same after a collision as before. The characteristics of the collision do not matter as long as it is an ideal system . In an ideal system, there is no friction or other real-world imperfection, and the total system momentum never changes unless a new mass or force is introduced.
The law of conservation of momentum applies not only to systems having two objects or particles but also to systems having any number of objects or particles. However, the law holds only in a closed system , that is, a system in which the total mass remains constant, and no forces are introduced from the outside.
This is a good time to make an important announcement. From now on in this book, if specific units are not given for quantities, assume that the units are intended to be expressed in the International System (SI). Therefore, in the following examples, masses are in kilograms, velocity magnitudes are in meters per second, and momentums are in kilogram-meters per second. Get into the habit of making this assumption, whether the units end up being important in the discussion or not. Of course, if other units are specified, then use those. But beware when making calculations. Units always must agree throughout a calculation, or you run the risk of getting a nonsensical or inaccurate result.
Sticky Objects
Look at Fig. 8-2. The two objects have masses m 1 and m 2 , and they are moving at speeds v 1 and v 2 , respectively. The velocity vectors v 1 and v 2 are not specifically shown here, but they point in the directions shown by the arrows. At A in this illustration, the two objects are on a collision course. The momentum of the object with mass m 1 is equal to p 1 = m 1 v 1 ; the momentum of the object with mass m 2 is equal to p 2 = m 2 v 2 .
Fig. 8-2 . ( a ) Two sticky objects, both with constant but different velocities, approach each other, ( b ) The objects after the collision.
At B , the objects have just hit each other and stuck together. After the collision, the composite object cruises along at a new velocity v that is different from either of the initial velocities. The new momentum, call it p , is equal to the sum of the original momentums. Therefore:
p = m 1 v 1 + m 2 v 2
The final velocity v can be determined by noting that the final mass is m 1 + m 2 . Therefore:
p = ( m 1 + m 2 ) v
v = p /( m 1 + m 2 )
Bouncy Objects
Now examine Fig. 8-3. The two objects have masses m 1 and m 2 , and they are moving at speeds v 1 and v 2 , respectively. The velocity vectors v 1 and v 2 are not specifically shown here, but they point in the directions shown by the arrows. At A in this illustration, the two objects are on a collision course. The momentum of the object with mass m 1 is equal to p 1 = m 1 v 1 ; the momentum of the object with mass m 2 is equal to p 2 = m 2 v 2 . Thus far the situation is just the same as that in Fig. 8-2. But here the objects are made of different stuff. They bounce off of each other when they collide.
Fig. 8-3 . ( a ) Two bouncy objects, both with constant but different velocities, approach each other, ( b ) The objects after the collision.
At B , the objects have just hit each other and bounced. Of course, their masses have not changed, but their velocities have, so their individual momentums have changed. However, the total momentum of the system has not changed, according to the law of conservation of momentum. Suppose that the new velocity of m 1 is v 1a and that the new velocity of m 2 is v 2a . The new momentums of the objects are therefore
p 1a = m 1 v 1a
p 2a = m 2 v 2a
According to the law of conservation of momentum,
p 1 + p 2 = p 1a + p 2a
and therefore:
m 1 v 1 + m 2 v 2 = m 1 v 1a + m 2 v 2a
The examples shown in Figs. 8-2 and 8-3 represent idealized situations. In the real world, there would be complications that we are ignoring here for the sake of demonstrating basic principles. For example, you might already be wondering whether or not the collisions shown in these drawings would impart spin to the composite mass (in Fig. 8-2) or to either or both masses (in Fig. 8-3). In the real world, this would usually happen, and it would make our calculations vastly more complicated. In these idealized examples, we assume that no spin is produced by the collisions.
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.
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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