Linear Momentum - Work, Energy, Power, and Momentum - Homework Helpers: Physics

Homework Helpers: Physics

3 Work, Energy, Power, and Momentum

Lesson 3–5: Linear Momentum

Watching a football game on television, you are likely to hear the sportscasters use the term momentum in a couple of different ways. Sometimes they talk about which team has momentum referring to which team seems to be controlling the game at the time. They might also talk about themomentum of a player carrying him forward. The second use of this term is closer to how we define linear momentum in physics. Linear momentum is the product of an object”s mass and velocity. We will use the symbol p for momentum, and the units are often shown as kg × m/s.

Linear momentum = mass × velocity

p = mv

Linear momentum (p) is measured in kg × m/s

For the rest of this lesson, we will simply refer to linear momentum as momentum, which is a common practice.

Momentum is a vector quantity, and the direction of the momentum will be in the same direction as the velocity. As with the case of velocity, the direction of the momentum can be indicated by a compass direction, or by a sign. A negative momentum simply means a momentum in the direction that is considered negative.

Example 1

Calculate the momentum of a 5.5 kg bowling ball traveling with a velocity of –15 m/s.

Given: m = 5.5 kg v = –15 m/s

Find: p

The momentum of an object will change anytime its velocity (or mass) changes. The tricky part about change in momentum problems is to pay attention to the direction of the velocity. If a ball hits a wall and bounces off at the exact same speed, its speed hasn”t changed, but its direction, and therefore its velocity, has. For questions that ask you to calculate the change in the momentum of a single object, where the mass of the object does not change, use the following formula.

Δp = pfpi = mvfmvi = m(vfvi)

Example 2

A baseball with a mass of 0.10 kg has a speed of 20.0 m/s as it flies toward a batter. The batter hits the ball and sends it in the exact opposite direction with a speed of 22.0 m/s. Calculate the change in the ball”s momentum.

The key to the problem is to notice that they give you the speed of the ball, not the velocity. If you read the question too fast, you might think that the velocity of the ball has only changed by 2.0 m/s, but that is not the case. Reading the problem more carefully, you find that the batter hit the ball and sent it in the exact opposite direction. That means that we need to make the final velocity −22.0 m/s, and the velocity of the ball actually changed by (–22.0 m/s – 20.0 m/s) –42.0 m/s!

Given: m = 0.10 kg vi = 20.0m/s vf = 22.0 m/s

Find: Δp

Solution: Δp = m(vfvi) = 0.10 kg(–22.0 m/s – 20.0 m/s)

 = 0.10 kg(–42.0 m/s) = –4.2 kg · m/s

In Lesson 3–2 we discussed the law of conservation of mechanical energy and said that there are cases in which the total mechanical energy of a system is conserved. There are also situations in which the total momentum of a system is conserved, resulting in the law of conservation of linear momentum.

Law of Conservation of Momentum

The total momentum of a system remains the same when the net external force acting on a system is zero.

General formula for two objects:

ΣΔp = (m1v1f + m2v2f) – (m1v1i + m2v2i) = 0

This formula is usually shown as m1v1i + m2v2i = m1v1f + m2v2f

The momentum of the baseball in Example 2 wasn”t conserved because it was acted on by an external force, which was provided by the bat. We could have studied both the bat and the ball as one system, and the law of conservation of momentum would have applied. As long as the air resistance and loss of energy to sound were treated as negligible, we would have found that the momentum gained by the ball was equal to the momentum lost by the bat.

A common external force that prevents momentum from being conserved in many real-life situations is friction. In a game of billiards, there is friction between the balls and the table, and between the balls and the air. If there were no external forces at work in a game of billiards, and if no energy was lost in collisions, every break shot, which sets all of the balls in motion, would likely result in each of the balls eventually being pocketed. The balls would simply bounce around on the table, colliding with each other, until they found a pocket.

An air hockey table comes much closer to approaching a frictionless surface. The puck floats on a cushion of air, and it will ricochet around on the table for a good amount of time. However, there is still a small amount of friction between the puck and the air, and a certain amount of energy is converted to sound when the puck hits the sides of the table.

Collisions come in many varieties. When a tennis ball hits a wall, it deforms momentarily, but it springs back into shape as is bounces off the wall. When a car hits a wall, the car deforms quite a bit, and it doesn”t spring back into shape, although it may “bounce” back from the wall. If a ball of clay strikes the same wall, it is likely to deform and stick to the wall.

Elastic Collisions

In an isolated elastic collision, both the momentum and the kinetic energy are conserved. In other words, the total momentum before the collision is equal to the total momentum after the collision and the total kinetic energy before the collision is equal to the total kinetic energy after the collision. There are no perfectly elastic collisions, but when the loss of kinetic energy is considered insignificant, the collision is still treated as elastic. If you studied the ideal gas laws in chemistry, you may recall that the collisions between molecules of gases are treated as if they are totally elastic. The collisions of a puck on the air hockey table would likely be considered elastic. The example of the tennis ball hitting a wall could be as well.

Many of the problems that you will encounter for momentum will take place on ice or an air table, and they should be treated as isolated systems, where momentum will be conserved. If the objects bounce off each other during collisions, they should be treated as elastic collisions.

One common example of a problem designed to test conservation of momentum involves two ice-skaters facing each other on a frictionless surface of ice. The skaters are initially at rest, so their total momentum before they push off (m1v1i + m2v2i = 0) is zero. They push off each other, and you are asked to use the mass of the skaters and the final velocity of one of the skaters to find the final velocity of the other. Remember: If the total momentum before and after the push is conserved (m1v1i + m2v2i = m1v1f + m2v2f), then the total momentum after the push (m1v1f + m2v2f = 0) must also be equal to zero.

Example 3

Two ice-skaters, with masses of 45 kg and 65 kg, respectively, are initially at rest and facing each other over a surface of ice. They push off from each other and the 45 kg skater moves away with a velocity of 6.0 m/s. Find the final velocity of the 65 kg skater.

The negative sign in our answer indicates that the 65 kg skater will move in the opposite direction of the 45 kg skater.

As you can see, when the objects start at rest, or when they end at rest, the formula for solving these equations is not very complicated. Let”s look at an example where the objects are moving before and after the collision.

Example 4

A ball with a mass of 0.25 kg and an initial velocity of 1.45 m/s has a head-on elastic collision with a 0.45 kg ball with an initial velocity of –2.25 m/s. After the collision, the 0.45 kg ball has a velocity of –0.84 m/s. Calculate the final velocity of the 0.25 kg ball. (Assume the momentum is conserved.)

Given: m1 = 0.25 kg m2 = 0.45 kg v1i = 1.45 m/s v2i = –2.25 m/s v2f = –0.84 m/s

Find: v1f

Solution: Our solution is based on the formula: m1v1i + m2v2i = m1v1f + m2v2f

Let”s start by calculating the total momentum before the collision:

m1v1i + m2v2i = (0.25 kg)(1.45 m/s) + (0.45 kg)(–2.25 m/s) = –0.65 kg· m/s

The law of conservation of momentum tells us that we must have the same momentum after the collision.

m1v1f + m2v2f = –0.65 kg· m/s

Isolating this formula for our unknown (v 1f), we get:

With these problems, I always feel that it is a good idea to check your answer. Let”s make sure that the total momentum was conserved by putting our numbers back into the original equation.

m1v1f + m2v2f = (0.25 kg)(–1.1 m/s) + (0.45 kg)(–0.84 m/s) = –0.65 kg · m/s

Our answer is correct, as momentum is conserved. The final velocity of the second ball is –1.1 m/s. It is interesting to note that the ball changed directions after being struck by the more massive, faster moving ball.

Inelastic Collisions

In a totally inelastic collision, the total kinetic energy of the system is not conserved. An inelastic collision is sometimes referred to as a “sticky” collision, because the objects remain in contact after the collision. Consider the following example.

Example 5

A block with a mass of 8.0 kg slides across a frictionless surface with an initial velocity of 5.0 m/s. It collides with a 4.0 kg block with a velocity of –10.0 m/s. After the collision, the two blocks stick together. Calculate the final velocity of the blocks.

Given: m1 = 8.0 kg m2 = 4.0 kg v1i = 5.0 m/s v2i = –10.0m/s

Find: vf

Solution: First, let”s find the initial total momentum of the two block system.

ptotal = m1v1i + m2v2i = (8.0 kg)(5.0 m/s) + (4.0 kg)(–10.0 m/s)

= 40. kg · m/s + (–40. kg · m/s) = 0.0 kg · m/s

The law of conservation of momentum tells us that the total initial momentum must equal the total final momentum, so the final momentum of the two-block system must also be equal to zero. The total mass of the blocks doesn”t change to zero just because they collide, so the velocity of the two blocks must equal zero.

You should see that the kinetic energy of this system is not conserved. Clearly the blocks had kinetic energy before they collided, but it is zero after the collision because both blocks end up at rest.

Example 6

A block with a mass of 6.0 kg and an initial velocity of 5.0 m/s collides with a stationary block with a mass of 4.0 kg on a frictionless surface. After the collision, the blocks stay together. What will be the final velocity of the two-block system?

Given: m1 = 6.0 kg m2 = 4.0 kg v1i = 5.0 m/s v2i = 0.0 m/s

Find: vf

Solution: The key to solving this problem is remembering that the blocks remain together after the collision, so that the final velocity of the blocks will be the same.

Momentum is conserved, so pi = pf

Original formula: m1v1i + m2v2i = m1v1f + m2v2f

The final velocity of both blocks is the same, so m1v1i + m2v2i = m1vf + m2vf

Factoring out vf: m1v1i + m2v2i = vf(m1 + m2)

Lesson 3–5 Review

1. A/an ________________ is a collision in which the total kinetic energy is conserved.

2. _________________ is the product of the mass and velocity of an object.

3. What is the velocity of a 0.10 kg baseball when its momentum is 3.50 kg × m/s?