Collision Example Problems

Collision Example Problems

Here’s our problem-solving strategy for conservation of momentum problems. We’ll use a set of boxes to help keep us organized and put all the information in the right place. The columns in our grid will help us compare the situation before and after the collision; we know that the total momentum before and after has to be the same. The rows of the grid will keep the objects separate in our analysis and allow us to easily add up the total momenta.

Example 1

A 10000 kg truck rolls down the road at 10 m/s when it runs headlong into a 1500 kg Volkswagen traveling at 20 m/s. The fenders of the two vehicles get tangled up and the vehicles move off together. With what speed and in what direction does the combined vehicle move off?

Solution: We know that momentum is always conserved, so we start by drawing some momentum boxes.

Remember:

• Only put momentum, not mass or velocity alone, into each box.
• Don’t combine objects’ masses in the boxes, even if they combine in the problem. For example, in this problem, the truck and car combine, but we still consider their masses separately when filling in the boxes in the “after” column.
• Add up the numbers down each column and put the total in the bottom row.
• Set the two “total” boxes equal to each other, then solve.

Q: But Mr. D, why did you put a positive number for the truck’s speed and a negative number for the car’s speed?

A: Because they’re moving in different directions, they have to have different signs. It doesn’t matter which one you choose to be negative, but one of them has to be. The fact that the answer comes out positive means that the combined wreckage is going in the direction that the truck was moving.

Q: But Mr. D, why did you use the same variable v for the speed of the truck and car after the collision?

A: When the problem tells you that two objects stick together or travel together or combine, or something like that, it’s your clue that they must have the same speed.

Example 2

A 500-kg cannon is at rest on its wheeled mount on a level, frictionless surface when it fires a 10-kg cannonball at 50 m/s. With what speed does the cannon recoil?

Solution: As usual, start with the boxes:

Q: But Mr. D, how did you know to put zeros in the “before” column?

A: Because the cannon and cannonball are both at rest before the ball is fired, we know the total momentum of both objects has to be zero. That means the total momentum of both objects after the collision also has to be zero; it also means that the individual momenta of the cannon and cannonball have to be equal but opposite. (NOT speeds, momenta.)

Q: But Mr. D, “momenta”?

A: It’s the plural of momentum. Not momentums, momenta.

Q: But Mr. D, the answer came out negative! Is that OK?

A: Because we called the 50 m/s for the cannonball positive, the negative velocity for the cannon means it moves in the opposite direction as the ball, which is exactly what we expect!