Lesson 2: The Law of Momentum ConservationUsing Equations as a Guide to ThinkingThe three problems on the previous page illustrate how the law of momentum conservation can be used to solve problems in which the after-collision velocity of an object is predicted based on mass-velocity information. There are similar practice problems (with accompanying solutions) lower on this page which are worth the practice. However, let's first take a more cognitive approach to some collision problems. The questions which follow provide a real test of your conceptual understanding of momentum conservation in collisions. Imagine that you are hovering next to a space shuttle in earth orbit and your buddy of equal mass who is moving at 4 km/hr (with respect to the ship) bumps into you. If she holds onto you, how fast do you move (with respect to the ship)? 
This problem could be solved in the
usual manner with a momentum table; the variable m could
be used for the mass of the astronauts or any random
number could be used for the mass of the astronauts
(provided each astronaut had the same mass). In the
process of solving the problem, the mass would cancel out
of the equation. However, there is a better, more
conceptual means of solving the problem. In order for the
momentum before the collision to be equal to the momentum
after the collision, the after collision velocity must be
smaller than the before collision velocity. 
The process of solving this problem involved using a conceptual understanding of the equation for momentum (p=m*v). This equation becomes a guide to thinking about how a change in one variable effects a change in another variable. The constant quantity in a collision is the momentum (momentum is conserved). For a constant momentum value, mass and velocity are inversely proportional. Thus, an increase in mass results in a decrease in velocity. 
A twofold increase in mass, results in a twofold decrease in velocity (the velocity is one-half its original value); a threefold increase in mass results in a threefold decrease in velocity (the velocity is one-third its original value); etc. Of course, it is instructive to point out that this form of problem-solving is limited to situations in which one of the two objects is at rest before the collision and both objects move at the same speed after the collision. To further test your understanding of this type of quantitative reasoning, try the following two questions. A large fish is in motion at 2 m/s when it encounters a smaller fish which is at rest. The large fish swallows the smaller fish and continues in motion at a reduced speed. If the large fish has three times the mass of the smaller fish, then what is the speed of the large fish (and the smaller fish) after the collision? Depress mouse on "pop-up menu" to view answer. 
A railroad diesel engine has four times the mass of a flatcar. If a diesel coasts at 5 km/hr into a flatcar that is initially at rest, how fast do the two coast if they couple together? Depress mouse on "pop-up menu" to view answer. 
Express your understanding of the concept and mathematics of momentum conservation by answering the following questions. Assume isolated systems and momentum conservation for each problem. Depress the mouse on the "pop-up" menu to view the answers. (If necessary, return to the instructional page on solving collision analysis problems.) 1. A 0.105-kg hockey puck moving at 48 m/s is caught by a 75-kg goalie at rest. With what velocity does the goalie slide on the ice after catching the puck? 2. A 35.0-g bullet strikes a 5.0-kg stationary wooden block and embeds itself in the block. The block and bullet move together at 8.6 m/s. What was the original velocity of the bullet? (CAUTION: Be careful of the units on velocity.) See Solution3. A 35.0-g bullet moving at 475 m/s strikes a 2.5-kg wooden block. The bullet passes through the block, leaving at 275 m/s. The block was at rest when it was hit. How fast is it moving when the bullet leaves? (CAUTION: Be careful of the units on velocity.) 4. A 0.50-kg ball traveling at 6.0 m/s collides head-on with a 1.00-kg ball moving in the opposite direction at a velocity of -12.0 m/s. The 0.50-kg ball moves away at -14 m/s after the collision. Find the velocity of the second ball. 5. A 3000-kg truck moving rightward with a speed of 5 km/hr collides head-on with a 1000-kg car moving leftward with a speed of 10 km/hr. The two vehicles stick together and move with the same velocity after the collision. Determine the post-collision velocity of the car and truck. (CAREFUL: Be cautious of the +/- sign on the velocity of the two vehicles.) 6. During a goal-line stand, a 75-kg fullback moving eastward with a speed of 8 m/s collides head-on with a 100-kg lineman moving westward with a speed of 4 m/s. The two players collide and stick together, moving at the same velocity after the collision. Determine the the post-collision velocity of the two players. (CAREFUL: Be cautious of the +/- sign on the velocity of the two players.) Answers 1. The problem can be solved using a momentum table: 
2. The problem can be solved using a momentum table:
3. The problem can be solved using a momentum table:
4. The problem can be solved using a momentum table:
5. The problem can be solved using a momentum table:
6. The problem can be solved using a momentum table:
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How
many times smaller must it be? By what factor must the
velocity be decreased? Before the collision, the amount
of mass in motion is m; after the collision, the amount
of mass in motion is 2*m. The amount of mass in motion
has doubled as the result of the collision. If the mass
is increased by a factor of two, then the velocity must
be decreased by a factor of 2. The before-collision
velocity was 4 km/hr so the after-collision velocity must
be one-half this value: 2 km/hr. Each astronaut is moving
with a velocity of 2 km/hr after the collision.
