Physics 12 Momentum Workbookname

Physics 12 Momentum WorkbookName:

Block:

Momentum and Impulse

1. Read and understand Ch 7.1 p. 149
2. This section describes the relationship between momentum and Newton's Second Law.

Momentum is given the symbol p and is defined as mass x velocity. Hence:

p = mv

Because momentum is the product of a scalar value, mass, and a vector value, velocity, it is a vector value itself. This means that direction of momentum is important.

In p = mv the only variable that can change for any one particular object is v Therefor, a change in momentum requires a change in velocity. Hence:

1. Do the following questions
2. A 900kg car’s momentum increases by 300kgm/s. What was the increase in velocity of the car?

1. A 300 g ball travelling east at 10 m/s rebounds off a wall and ends up travelling west at 7 m/s. What was the change in momentum of the ball?

1. Clearly, to change an object's velocity you would need to apply a force over a period of time. Look at the following sequence of equations and make sure you can understand each step starting from the left and reading to the right. Actually, it does not matter which side you read from.

So,

Note that is a force applied over a period of time; thus, causing a change in momentum. Like this:

is called an impulse and is the same as the change in momentum.

An impulse causes a change in momentum.So we have

Conservation of Momentum in One Dimension

1. Read and understand Ch. 7.2 p. 151 Conservation of Momentum

The total momentum before a collision is the same as the total momentum after the collision.

There are three main situations where the above equation can be applied. Do the example for each case. Hints are given along the way.

Case 1 A simple collision where the objects separate after the collision.

A 1kg ball travelling west at 3m/s strikes another ball of 4kg also travelling west but at 2m/s. After the collision the 4kg ball ends up travelling west at 2.5m/s. What is the velocity of the 1kg ball? Hint: list the information and determine which variable of the above equation you need to solve for.

Case 2 A collision where the objects stick together.

A 5g bullet travelling at 250m/s strikes a stationary 1.5 kg block of wood. The bullet becomes imbedded and the two move off together. After the collision, what is the velocity of the block and bullet together? Hint: Since the two objects are stuck together they have the same velocity after the impact. Therefor, and they can be factored in the right hand side of . Like this: .

So, Now it is easy to solve for

Case 3 Explosions.

Two carts initially touching are pushed apart by the explosion of a firecracker. After the explosion object 1, mass 3kg, moves west at 1 m/s . What is the velocity of object 2 after the explosion?

p0 = p1 + p2

1. Do Energy and Momentum #1

1. Do Problems 1 – 10, 13 – 17 on page

Conservation of Momentum in Two Dimensions

1. There are two things to remember when you are working on momentum problems in two dimensions.
2. During a collision the momentum of each object is conserved in the direction in which it was originally going.
3. You can break up a momentum vector into its x and y components.

1. In the above diagram ball 1 has a mass of 700g and is moving at 2m/s [800 S of E] and ball 2 has a mass of 300 g and is traveling at 3m/s [200 N of E]. They collide ball 1 ends up traveling at 0.5 m/s [350 N of E] and ball two heads off in an unknown direction roughly SW.

1. Find the x and y components of each ball’s momenta before the collision.

1. Find the x and y components of ball 1 after the collision.

1. Solve for the x and y components for the unknown momentum by completing the following matrix. p1x + p2x = p1x’ + p2x’ and p1y + p2y = p1y’ + p2y’

XY

p1 = (,)

+p2 = (,)

-p1’ = (,)

p2’ = (,)

1. Momentum and Energy ExerciseName: Blk:

1. Momentum is always conserved. It does not matter whether there is energy lost or not momentum is always conserved.
2. Energy is not always conserved as kinetic energy. Most of the time energy is lost due to thermal energy loss as a result of friction and deformation of material, and light. This kind of collision is said to be inelastic.
3. When there is a collision that does not loose kinetic energy then the collision is said to be elastic.

One of the following examples is elastic and the other is inelastic. You job is to figure out which is the elastic collision. Remember that energy is NOT a vector. For energy, direction is not important. Therefore, you can simply add the kinetic energy of the objects before the collision and compare it to the kinetic energy of the objects after the collision.

For each diagram use a scale of 1cm=1m and a time interval of ∆t=0.1s.

Collision 1

m1=4kg

m2=3kg
Collision 2

m1 = m2 = 5kg

v2 = 0m/s

v1’ is perpendicular to v2’

Which was the elastic collision?

What must happen for a collision to be elastic?

Using Sine Law and Cosine Law for Momentum in 2D

1. When you did the Momentum Exercise you used components to work out the momenta of the objects after the collision. You may often find it much more convenient to use either Sine Law or Cosine Law.
2. Three steps to solving using trig. Functions.
3. Draw a vector diagram showing the two initial momenta of the colliding objects and their resultant vector. v1 + v2 = vR
4. Suppose you are trying to find v2’; then, redraw the original vR and sketch in v1’. This method works because

Note that : v1’ + v2’ = vR As a result, the v2’can be sketched in to complete the triangle.

1. Use trigonometry to find the values.

1. Two balls of masses m1=0.20 kg and m2=0.15 kg collide. m1 was originally travelling at 0.76 m/s at 10 degrees down from the positive x axis and ended up travelling at 0.30 m/s at 37 degrees up from the positive x axis. m2 was originally travelling at 0.50 m/s at 26 degrees up from the positive x-axis. What is the velocity and direction of m2 after the collision?

1. Suppose two balls of equal mass collided. M1 was traveling at 0.6 m/s parallel to the positive x axis and after the collision was traveling at 0.8 m/s at 20 degrees up from the positive x axis. After the collision M2 was traveling at 0.55 m/s at 10 degrees down from the positive x-axis. What was the original velocity of M2?

1. Do Questions 25 - 29 p. 169 – 170

1. Do Questions 52, 54, 56, 57, 58, 61

1. Do Momentum Work Sheet.

Tolksdorff Momentum Workbook1 of 7