Acceleration WorksheetName: ____________________Period: _______

1. A turtle has a speed of 0.50 m/s. After 6 seconds, it has a speed of 0.80 m/s. What is the turtles average acceleration in m/s? (0.05)

2. What is a sports car average acceleration if it can go from 0 m/s to 27 m/s in 6.0 sec? (4.5)

3. A school bus can accelerate from a complete stop at 1.3 m/s2. How long will it take the bus to reach a speed of 12.1 m/s? (9.3)

4. If a bicycle has an average acceleration of 0.44 m/s2, and its initial forward velocity is 8.2 m/s, what is its final velocity after 10 seconds? (12.6)

5. What is a cars acceleration when there is an accident on the highway the car slows down from an initial velocity of 24.5 m/s to a final velocity of 4.5 m/s in 3.2 seconds. Is it acceleration or deceleration? (-6.25)

Average Speed and Average Velocity Problems:

1. A car travels west 50 km for 2 hours before heading south at a rate of 30 kilometers per hour for 1.5 hours. What is the cars average speed and average velocity? Dont forget directions on velocity.

2. A swimmer competes in the 500-meter free. Their swim time is 11 minutes and 32 seconds. What is their average speed in meters per second? What is their average velocity in km/hr?

3. A hiker has an average velocity of 5 km/hr NE for 3 hours. The hiker traveled a distance of 10 meters east and X meters north. What distance did the hiker travel to the north?

4. What was the hikers average speed in #3?

5. A cyclist travels east for 3 hours and covers 3 km before heading north for an additional 2 hours. The cyclists displacement is in the direction of 30 degrees Northeast. What is the cyclists average speed and average velocity?

Momentum = mass velocityIn physics, the symbol for the quantity momentum is the lower case "p". Thus, the above equation can be rewritten asp = m vwheremis the mass andvis the velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity.The units for momentum would be mass units times velocity units. The standard metric unit of momentum is the kgm/s. While the kgm/s is the standard metric unit of momentum, there are a variety of other units that are acceptable (though not conventional) units of momentum. Examples include kgmi/hr, kgkm/hr, and gcm/s. In each of these examples, a mass unit is multiplied by a velocity unit to provide a momentum unit. This is consistent with the equation for momentum.Momentum is avector quantity. As discussed in an earlier unit,a vector quantityis a quantity that is fully described by both magnitude and direction. To fully describe the momentum of a 5-kg bowling ball moving westward at 2 m/s, youmust include information about both the magnitude and the direction of the bowling ball. It isnotenough to say that the ball has 10 kgm/s of momentum; the momentum of the ball isnotfully described until information about its direction is given. The direction of the momentum vector is the same as the direction of the velocity of the ball. In a previous unit, it was said thatthe direction of the velocity vectoris the same as the direction that an object is moving. If the bowling ball is moving westward, then its momentum can be fully described by saying that it is 10 kgm/s, westward. As a vector quantity, the momentum of an object is fully described by both magnitudeand direction.From the definition of momentum, it becomes obvious that an object has a large momentum if either its mass or its velocity is large. Both variables are of equal importance in determining the momentum of an object. Consider a Mack truck and a roller skate moving down the street at the same speed. The considerably greater mass of the Mack truck gives it a considerably greater momentum. Yet if the Mack truck were at rest, then the momentum of the least massive roller skate would be the greatest. The momentum of any object that is at rest is 0. Objects at rest donothave momentum - they do not have any "mass in motion." Both variables - mass and velocity - are important in comparing the momentum of two objects.Themomentum equationcan help us to think about how a change in one of the two variables might affect the momentum of an object. Consider a 0.5-kg physics cart loaded with one 0.5-kg brick and moving with a speed of 2.0 m/s. The total mass ofloadedcart is 1.0 kg and its momentum is 2.0 kgm/s. If the cart was instead loaded with three 0.5-kg bricks, then the total mass of theloadedcart would be 2.0 kg and its momentum would be 4.0 kgm/s. A doubling of the mass results in a doubling of the momentum.Similarly, if the 2.0-kg cart had a velocity of 8.0 m/s (instead of 2.0 m/s), then the cart would have a momentum of 16.0 kgm/s (instead of 4.0 kgm/s). Aquadruplingin velocity results in a quadruplingof the momentum. These two examples illustrate how the equation p = mv serves as a "guide to thinking" andnot merelya "plug-and-chugrecipe for algebraic problem-solving."Check Your UnderstandingExpress your understanding of the concept and mathematics of momentum by answering the following questions. Click the button to view the answers.1. Determine the momentum of a ...a. 60-kg halfback moving eastward at 9 m/s.

b. 1000-kg car moving northward at 20 m/s.

c. 40-kg freshman moving southward at 2 m/s.2. A car possesses 20 000 units of momentum. What would be the car's new momentum if ...a. its velocity was doubled.

b. its velocity was tripled.

c. its mass was doubled (by adding more passengers and a greater load)

d. both its velocity was doubled and its mass was doubled.

3. A halfback (m = 60 kg), a tight end (m = 90 kg), and a lineman (m = 120 kg) are running down the football field. Consider theirticker tape patternsbelow.

Compare the velocities of these three players. How many times greater are the velocity of the halfback and the velocity of the tight end than the velocity of the lineman? Which player has the greatest momentum? Explain.

Determining the Slope on a p-t GraphThe slope of the line on a position versus time graph is equal to the velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s. If the object is moving with a velocity of -8 m/s, then the slope of the line will be -8 m/s. If the object has a velocity of 0 m/s, then the slope of the line will be 0 m/s. The slope of the line on a position versus time graph tells it all. Because of its importance, a student of physics must have a good understanding of how to calculate the slope of a line. In this part of the lesson, the method for determining the slope of a line on a position-time graph will be discussed.Let's begin by considering the position versus time graph below.

The line is sloping upwards to the right. But mathematically, by how much does it slope upwards for every 1 second along the horizontal (time) axis? To answer this question we must use the slope equation.

The slope equation says that the slope of a line is found by determining the amount of rise of the line between any two points divided by the amount of run of the line between the same two points. In other words, Pick two points on the line and determine their coordinates. Determine the difference in y-coordinates of these two points (rise). Determine the difference in x-coordinates for these two points (run). Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).The diagram below shows this method being applied to determine the slope of the line. Note that three different calculations are performed for three different sets of two points on the line. In each case, the result is the same: the slope is 10 m/s.

So that was easy - rise over run is all that is involved.Now let's attempt a more difficult example. Consider the graph below. Note that the slope is not positive but rather negative; that is, the line slopes in the downward direction. Note also that the line on the graph does not pass through the origin. Slope calculations are relatively easy when the line passes through the origin since one of the points is (0,0). But that is not the case here. Test your understanding of slope calculations by determining the slope of the line below. Then click the button to check your answer.

Check Your Understanding1. Determine the velocity (i.e., slope) of the object as portrayed by the graph below.

Determining the Slope on a v-t GraphThe slope of the line on a velocity versus time graph is equal to theacceleration of the object. If the object is moving with an acceleration of +4 m/s/s (i.e., changing its velocity by 4 m/s per second), then the slope of the line will be +4 m/s/s. If the object is moving with an acceleration of -8 m/s/s, then the slope of the line will be -8 m/s/s. If the object has a velocity of 0 m/s, then the slope of the line will be 0 m/s. Because of its importance, a student of physics must have a good understanding of how to calculate the slope of a line. In this part of the lesson, the method for determining the slope of a line on a velocity-time graph will be discussed.

Let's begin by considering the velocity versus time graph below.

The line is sloping upwards to the right. But mathematically, by how much does it slope upwards for every 1 second along the horizontal (time) axis? To answer this question we must use the slope equation.

The slope equation says that the slope of a line is found by determining the amount ofriseof the line between any two points divided by the amount ofrunof the line between the same two points. A method for carrying out the calculation isa. Pick two points on the line and determine their coordinates.b. Determine the difference in y-coordinates for these two points (rise).c. Determine the difference in x-coordinates for these two points (run).d. Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).

The diagram below shows this method being applied to determine the slope of the line. Note that three different calculations are performed for three different sets of two points on the line. In each case, the result is the same: the slope is 10 m/s/s.

Observe that regardless of which two points on the line are chosen for the slope calculation, the result remains the same - 10 m/s/s.Check Your UnderstandingConsider the velocity-time graph below. Determine the acceleration (i.e., slope) of the object as portrayed by the graph.

Using Equations as a Recipe for Algebraic Problem-SolvingThe momentum lost by one object is equal to the momentum gained by another object. For collisions occurring in anisolated system, there are no exceptions to this law. This law becomes a powerful tool in physics because it allows for predictions of the before- and after-collision velocities (or mass) of an object. In this portion of Lesson 2, the law of momentum conservation will be used to make such predictions. The law of momentum conservation will be combined with the use of a "momentum table" and some algebra skills to solve problems involving collisions occurring in isolated systems.Consider the following problem:A 15-kg medicine ball is thrown at a velocity of 20 km/hr to a 60-kg person who is at rest on ice. The person catches the ball and subsequently slides with the ball across the ice. Determine the velocity of the person and the ball after the collision.Such a motion can be considered as a collision between a person and a medicine ball. Before the collision, the ball has momentum and the person does not. The collision causes the ball to lose momentum and the person to gain momentum. After the collision, the ball and the person travel with the same velocity (v) across the ice.

If it can be assumed that the affect of friction between the person and the ice is negligible, then the collision has occurred in anisolated system. Momentum should be conserved and the post-collision velocity (v) can be determined using a momentum table as shown below.Before CollisionAfter Collision

Person0(60 kg) v

Medicine ball(15 kg) (20 km/hr)= 300 kg km/hr(15 kg) v

Total300 kg km/hr300

To determinev(the velocity of both the objects after the collision), the sum of the individual momentum of the two objects can be set equal to the total system momentum. The following equation results:60 v + 15 v = 30075 v = 300v = 4 km/hr

Now consider a similar problem involving momentum conservation.A 0.150-kg baseball moving at a speed of 45.0 m/s crosses the plate and strikes the 0.250-kg catcher's mitt (originally at rest). The catcher's mitt immediately recoils backwards (at the same speed as the ball) before the catcher applies an external force to stop its momentum. If the catcher's hand is in a relaxed state at the time of the collision, it can be assumed that no net external force exists and the law of momentum conservation applies to the baseball-catcher's mitt collision. Determine the post-collision velocity of the mitt and ball.Before the collision, the ball has momentum and the catcher's mitt does not. The collision causes the ball to lose momentum and the catcher's mitt to gain momentum. After the collision, the ball and the mitt move with the same velocity (v).

The collision between the ball and the catcher's mitt occurs in anisolated system,total system momentum is conserved. Thus, the total momentum before the collision (possessed solely by the baseball) equals the total momentum after the collision (shared by the baseball and the catcher's mitt). The table below depicts this principle of momentum conservation.Before CollisionAfter Collision

Ball0.15 kg 45 m/s = 6.75 kgm/s(0.15 kg) v

Catcher's Mitt0(0.25 kg) v

Total6.75 kgm/s6.75 kgm/s

To determinev(the velocity of both objects after the collision), the sum of the individual momentum of the two objects is set equal to the total system momentum. The following equation results:0.15 kg v + 0.25 kg v =6.75 kgm/s0.40 kg v =6.75 kgm/sv = 16.9 m/sUsing algebra skills, it can be shown that v = 16.9 m/s. Both the baseball and the catcher's mitt move with a velocity of 16.9 m/s immediately after the collision and prior to the moment that the catcher begins to apply an external force.The two collisions above are examples of inelastic collisions.Now we will consider the analysis of a collision in which the two objects donotstick together. In this collision, the two objects willbounce offeach other. While this is not technically an elastic collision, it is more elastic than the previous collisions in which the two objectsstick together.A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 15 m/s. Assuming that momentum is conserved during the collision, determine the velocity of the truck immediately after the collision.In this collision, the truck has a considerable amount of momentum before the collision and the car has no momentum (it is at rest). After the collision, the truck slows down (loses momentum) and the car speeds up (gains momentum).

The collision can be analyzed using a momentum table similar to the above situations.Before CollisionAfter Collision

Truck3000 10 = 30 0003000 v

Car01000 15 = 15 000

Total30 00030 000

To determinev(the velocity of the truck), the sum of the individual after-collision momentum of the two objects is set equal to the total momentum. The following equation results:3000*v + 15 000 = 30 0003000*v = 15 000v = 5.0 m/sThe next section of this lesson involves examples of problems that provide arealtest of your conceptual understanding of momentum conservation in collisions. Before proceeding with thepractice problems, be sure to try a few of the moreconceptual questions that follow.Practices.a. An airplane accelerates down a runway at 3.20 m/s2for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

b. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

c. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity and how far will he fall?

d. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.

e. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.

f. Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.8 seconds, then what is the acceleration and what is the distance that the sled travels?g. A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.

h. An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?

i. A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration).

j. A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.

k. If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?

l. A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).

m. A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the peak is one-half the total hang-time.)

The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.

n. A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.)

o. A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.

p. It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it began to skid.

q. A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.

r. A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the acceleration (assume uniform) of the dragster.

s. With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.

Practice Problems1) A block of aluminum occupies a volume of 15.0 mL and weighs 40.5 g. What is its density?

2) Mercury metal is poured into a graduated cylinder that holds exactly 22.5 mL. The mercury used to fill the cylinder weighs 306.0 g. From this information, calculate the density of mercury.

3) What is the weight of the ethyl alcohol that exactly fills a 200.0 mL container? The density of ethyl alcohol is 0.789 g/mL.

4) A rectangular block of copper metal weighs 1896 g. The dimensions of the block are 8.4 cm by 5.5 cm by 4.6 cm. From this data, what is the density of copper?

5) A flask that weighs 345.8 g is filled with 225 mL of carbon tetrachloride. The weight of the flask and carbon tetrachloride is found to be 703.55 g. From this information, calculate the density of carbon tetrachloride.

6) Calculate the density of sulfuric acid if 35.4 mL of the acid weighs 65.14 g.

7) Find the mass of 250.0 mL of benzene. The density of benzene is 0.8786 g/mL.

8) A block of lead has dimensions of 4.50 cm by 5.20 cm by 6.00 cm. The block weighs 1591 g. From this information, calculate the density of lead.

9) 28.5 g of iron shot is added to a graduated cylinder containing 45.5 mL of water. The water level rises to the 49.1 mL mark, From this information, calculate the density of iron.

10) What volume of silver metal will weigh exactly 2500.0 g. The density of silver is 10.5 g/cm3.