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SPEED and ACCELERATION Speed is used to measure how fast an object moves. It is calculated by the distance an object travels divided by the time it takes the object to travel that distance. The formula for speed is s = d/t, where s represents speed, d represents distance, and t represents time. In the United States, speed is usually measured in miles per hour (mph) or feet per second (ft/s.) The “/” symbol represents the word “per.” For scientifi c purposes, we will use metric units for distance: meters (m) and kilometers (km.) For perspective, 1 m/s = 2.24 mph, and 1 km/hr = 0.62 mph. Velocity is measured in the same units as speed, but the direction of motion is also given. Direction may be described, for example, as north, south, east, west, left, or right. Velocity is said to be a vector quantity, meaning both magnitude (distance from one point to another) and direction are specifi ed. Speed is said to be a scalar quantity, meaning it is a number with no indication of direction. An arrow on a grid often represents a vector quantity. The length of the arrow indicates the magnitude, and the point of the arrow shows the direction; the vector on the graph on the right shows that a car is traveling northeast at 80 km/hr. Acceleration is the rate of velocity change during a certain period of time. Since the unit for velocity has time in the denominator (m/s for example), and velocity is divided by units of time (s for example) to calculate acceleration, the unit of time is given in the denominator twice. So, the units for acceleration are m/s2 , which is stated as “meters per second per second” or “meters per second squared.” In everyday usage, acceleration is usually thought of as change in speed. However, it is important to remember that acceleration is change in velocity. So, even if an object is moving at a constant speed, if it changes direction, the object is accelerating. The speed of an object is the rate at which it covers distance. The general formula for speed is distance divided by time. We write speed = distance/time, v = d/t. For the sake of convenience we use the abbreviations d = distance, t = time, and v = speed. If you cover a distance of 80 miles in two hours, then your average speed is v = (80 miles)/(2 hours) = 40 miles/hour. You probably go slower than average during some periods of time, and faster during other periods. The speedometer of the car shows the car's i nstantaneous speed, i.e. how fast it is going at any moment. Link: Average vs. Instantaneous Speed
Transcript

SPEED and ACCELERATION

Speed is used to measure how fast an object moves. It is calculated by the distance an object travels divided by the time it takes the object to travel that distance. The formula for speed is s = d/t, where s represents speed, d represents distance, and t represents time. In the United States, speed is usually measured in miles per hour (mph) or feet per second (ft/s.) The “/” symbol represents the word “per.” For scientifi c purposes, we will use metric units for distance: meters (m) and kilometers (km.) For perspective, 1 m/s = 2.24 mph, and 1 km/hr = 0.62 mph.

Velocity is measured in the same units as speed, but the direction of motion is also given. Direction may be described, for example, as north, south, east, west, left, or right. Velocity is said to be a vector quantity, meaning both magnitude (distance from one point to another) and direction are specifi ed. Speed is said to be a scalar quantity, meaning it is a number with no indication of direction. An arrow ona grid often represents a vector quantity. The length of the arrow indicates the magnitude, and the point of the arrow shows the direction; the vector on the graph on the right shows that a car is traveling northeast at 80 km/hr.

Acceleration is the rate of velocity change during a certain period of time. Since the unit for velocity has time in the denominator (m/s for example), and velocity is divided by units of time (s for example) to calculate acceleration, the unit of time is given in the denominator twice. So, the units for accelerationare m/s2 , which is stated as “meters per second per second” or “meters per second squared.” In everyday usage, acceleration is usually thought of as change in speed. However, it is important to remember that acceleration is change in velocity. So, even if an object is moving at a constant speed, if itchanges direction, the object is accelerating.The speed of an object is the rate at which it covers distance. The general formula for speed is distance divided by time. We write

speed = distance/time, v = d/t.

For the sake of convenience we use the abbreviations d = distance, t = time, and v = speed. If you cover a distance of 80 miles in two hours, then your average speed is

v = (80 miles)/(2 hours) = 40 miles/hour.

You probably go slower than average during some periods of time, and faster during other periods. The speedometer of the car shows the car's instantaneous speed, i.e. how fast it is going at any moment.

Link:

Average vs. Instantaneous Speed

According to the rules of algebra, we can rewrite the formula v = d/t in two different ways.

If we want to know how far a car going at 55 miles/hour travels in 3hours we write d = vt = (55 miles/hour)*(3 hours) = 165 miles.

If we want to know how long it will take this car to cover a distance of 220 miles we write t = d/v = (220 miles)/(55 miles/hour) = 4 hours.

The units, such as miles and hours are always carried along in calculations and are treated like ordinary algebraic quantities.

The result of every measurement has two parts, a number and a unit. The number is the answer to "How many?" and the unit is the answer to "Of what". Unitsare standard quantities such as a second, a meter, a mile. The most widely used units today are those of the international system, abbreviated SI (Système International d'Unitès). Examples of SI units are the meter (m) for length, the second (s) for time, and thekilogram (kg) for mass.

The speed of an object is a scalar quantity. It just tells us how fast the object is moving, but not in which direction it is headed. The vector quantity which specifies both the speed and the direction is called the velocity. (Notation: speed = v, velocity = v.)

Turza’s Physical Science Semester Exam Review – Part TwoForces and Motion: Speed, Velocity, and Acceleration

EQUATIONS:Speed:     Velocity: Acceleration: Force:

Sample Problems:

A girl travels 20 miles on her bicycle. The trip takes 2 hours. Express her speed in miles/hr.

1. First, we identify the variables in our problem:

distance (d) = 20 miles time (t) = 2 hours

2. We place the variables in their correct position in the speed formula

S = d/t S = 20 mi/2 hour

3. Perform the calculation and express the resulting speed value with the appropriate unit:

S = 10 mi/hr

A car starts from a stoplight and is traveling with a velocity of 10 m/sec east in 20 seconds. What is the acceleration of the car?

1. First we identify the information that we are given in the problem:

vf - 10 m/sec vo - 0 m/sec time - 20 seconds

2. Then we insert the given information into the acceleration formula:

a = (vf - vo)/t a = (10 m/sec - 0 m/sec)/20 sec

3. Solving the problem gives an acceleration value of 0.5 m/sec2.

Now try on your own:1. What is the speed of a rocket that travels 9000 meters in 12.12 seconds? 742.57 m/s

2. What is the speed of a jet plane that travels 528 meters in 4 seconds? 132 m/s

3. How long will your trip take (in hours) if you travel 350 km at an average speed of 80 km/hr? 4.38 h

4. How far (in meters) will you travel in 3 minutes running at a rate of 6 m/s? 1,080 m

5. A trip to Cape Canaveral, Florida takes 10 hours. The distance is 816 km. Calculate the average speed. 81.6 km/h

6. How many seconds will it take for a satellite to travel 450 km at a rate of 120 m/s? 3,750 s

7. What is the speed of a walking person in m/s if the person travels 1000 m in 20 minutes? 0.80 m/s

8. A ball rolls down a ramp for 15 seconds. If the initial velocity of the ball was 0.8 m/sec and the final velocity was 7 m/sec, what was the acceleration of the ball ? 0.413 m/s²

9. A meteoroid changed velocity from 1.0 km/s to 1.8 km/s in 0.03 seconds. What is the acceleration of the meteoroid? 26.7 km/ s²

10. A car going 50mph accelerates to pass a truck. Five seconds later the car is going 80mph. Calculate the acceleration of the car. 6 mph/s

11. The space shuttle releases a space telescope into orbit around the earth. The telescope goes from being stationary to traveling at a speed of 1700 m/s in 25 seconds. What is the acceleration of the satellite? 68 m/s²

12. A ball is rolled at a velocity of 12 m/sec. After 36 seconds, it comes to a stop. What is the acceleration of the ball? -0.33 m/s²

13. How much force is needed to accelerate a truck with a mass of 2,000 kg, at a rate of 3 m/s²? 6,000 N

14. A dragster in a race accelerated from stop to 60 m/s by the time it reached the finish line. The dragster moved in a straight line and traveled from the starting line to the finish line in 8.0 sec. What was the acceleration of the dragster? 7.5 m/s²

15. A 300 N force acts on a 25 kg object. The acceleration of the object is 12 m/s²

Question 31

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Question: A certain force exerted for 1.2 seconds raises the speed of an object from 1.8 m/s to 4.2 m/s. Later

the same force is applied for 2 seconds. How much does the velocity change in 2 seconds?

Answer: Initial velocity (u) = 1.8 m/s.

Final velocity (v) = 4.2 m/s

Time (t) = 1.2 seconds

First calculate acceleration

As the same force acts for the next two seconds the acceleration produced will be the same. The final velocity in

the first case will now become the initial velocity. We have to calculate the final velocity at end of 2 seconds.

Acceleration (a) = 2 m/s2

Intial velocity (u) = 4.2 m/s, t = 2 s

Final velocity (v) = ?

[First equation of motion]

Change in velocity in two seconds = 8.2 - 4.2 = 4 m/s.

Question 32

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Question: Two blocks made of different metals, identical in shape and size are acted upon by equal forces,

which cause them to slide on a horizontal surface. The acceleration of the second block is found to be 5 times that

of the first.What is the ratio of the mass of second block to that of the first.

Answer: Let m1 and m2 be the mass of the first and second block respectively and a be the acceleration.

Given :

Question 33

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Question: Brakes are applied on a car of mass 1000 kg moving with a velocity of 54 km/h. The car covers a

distance of 50 m before coming to rest. Calculate the force applied on the car.

Answer: Initial velocity u = 54 km/h

= 15 m/s

Final velocity v = 0 (because the car is brought to rest)

Distance covered S = 50 m

We have to calculate force applied, i.e., F = ma. But we do not know the value of a. Making use of III equation of

motion we get,

F = ma

Mass = 1000 kg

Force = 1000 x (- 2.25)

= - 2250 N

Retarding force = 2250 N

Question 34

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Question: A car of mass 800 kg, moving at 108 km/h is brought to rest over a distance of 15 m. Find the

retarding force developed by the brakes of the car.

Answer: Initial velocity (u) = 108 km/hr

= 30 m/s

Final velocity (v) = 0

Distance covered (S) = 15 m

[III equation of motion] Applying v2 - u2 = 2aS

Mass of the car = 800 kg

Retarding force = 24000 N

Question 35

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Question: The velocity of a car of mass 1800 kg increases by 90 km/h in 15 seconds. Calculate the force

applied to increase the speed.

Answer: Mass of the car (m) = 1800 kg

= 5 x 5

= 25 m/s

Force applied = ma

Time (t) = 15 seconds

Force = ma

= 600 x 5

= 3000 N

Question 36

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Question: What is the acceleration produced by a force of 12 Newton exerted on an object of mass 3 kg?

Answer: From Newton's second law of motion F = ma

F = 12 N, mass (m) = 3 kg

Question 37

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Question: What force would be needed to produce an acceleration of 4 m/s2 on a ball of mass 6 kg?

Answer: Acceleration (a) = 4 m/s2

Mass of the ball (m) = 6 kg

Question 38

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Question: Why does a fielder in a cricket match move his hand backwards while making an attempt to catch the

ball?

Answer: By moving the hand backwards the fielder increases the time of action of the force on his hands.

When the time increases the force is reduced and the chances of the fielder getting hurt and dropping the ball is

less.

Question 39

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Question: For how long should a force of 200 N act on a body of 20 kg, so that it acquires a velocity of 100 m/s?

Answer: Time required to change the velocity of the body is to be calculated.

Initial Velocity (u) = 0

Final Velocity (v) = 100 m/s

The force should act on the body for 10 seconds.

Question 40

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Question: A retarding force is applied for 10 seconds to stop 2000 kg heavy car moving with a uniform velocity

of 40 m/s. Calculate the retardation and the retarding force.

Answer: Time (t) = 10 seconds

Final Velocity (v) = 0 [as the car comes to rest]

Initial velocity (u) = 40 m/s

Mass of the car = 2000 kg

Force required to stop the car = ma

= 2000 x -4

= -8000 N

Retarding Force = +8000 N

PROJECTILE

A projectile is any object that is cast, fired, flung, heaved, hurled, pitched, tossed, or thrown. (This is an informal definition.) The path of a projectile is called its trajectory. Some examples of projectiles include…

a baseball that has been pitched, batted, or throwna bullet the instant it exits the barrel of a gun or riflea bus driven off an uncompleted bridgea moving airplane in the air with its engines and wings disableda runner in mid stride (since they momentarily lose contact with the ground)the space shuttle or any other spacecraft after main engine cut off (MECO)

A projectile is an object upon which the only force acting is gravity. There are a variety of examples of projectiles. An object dropped from rest is a projectile (provided that the influence of air resistance is negligible). An object that is thrown vertically upward is also a projectile (provided that the influence of airresistance is negligible). And an object which is thrown upward at an angle to the horizontal is also a projectile (provided that the influence of air resistance is negligible). A projectile is any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.

Sample problems

Explore More

1. A stone is thrown horizontally at a speed of 8.0m/s from the edge of a cliff 80m in

height. How far from the base of the cliff will the stone strike the ground?

2. A toy truck moves off the edge of a table that is 1.25m high and lands 0.40m from the

base of the table.a. How much time passed between the moment the car left the table and the

moment it hit the floor?b. What was the horizontal velocity of the car when it hit the ground?

3. A hawk in level flight 135m above the ground drops the fish it caught. If the hawk’s

horizontal speed is 20.0m/s, how far ahead of the drop point will the fish land?

4. A pistol is fired horizontally toward a target 120m away, but at the same height. The

bullet’s velocity is 200m/s. How long does it take the bullet to get to the target? How far

below the target does the bullet hit?

5. A bird, traveling at 20m/s, wants to hit a waiter 10m below with his dropping (see

image). In order to hit the waiter, the bird must release his dropping some distance before

he is directly overhead. What is this distance? 6. Joe Nedney of the San Francisco 49ers kicked a field goal with an initial velocity

of 20m/s at an angle of 60∘.

a. How long is the ball in the air? Hint: you may assume that the ball lands at same height as it starts at.

b. What are the range and maximum height of the ball?

7. A racquetball thrown from the ground at an angle of 45∘ and with a speed

of 22.5m/slands exactly 2.5s later on the top of a nearby building. Calculate the

horizontal distanceit traveled and the height of the building.

8. Donovan McNabb throws a football. He throws it with an initial velocity of 30m/s at an

angle of 25∘. How much time passes until the ball travels 35m horizontally? What is the

height of the ball after 0.5 seconds? (Assume that, when thrown, the ball is 2m above the

ground.)

9. Pablo Sandoval throws a baseball with a horizontal component of velocity of 25m/s.

After 2 seconds, the ball is 40m above the release point. Calculate the horizontaldistance it

has traveled by this time, its initial vertical component of velocity, and its initial angle of projection. Also, is the ball on the way up or the way down at this moment in time?

10. Barry Bonds hits a 125m(450′) home run that lands in the stands at an

altitude 30mabove its starting altitude. Assuming that the ball left the bat at an angle

of 45∘ from the horizontal, calculate how long the ball was in the air.

11. A golfer can drive a ball with an initial speed of 40.0m/s. If the tee and the green are

separated by 100m, but are on the same level, at what angle should the ball be driven?

(Hint: you should use 2cos(x)sin(x)=sin(2x) at some point.)

12. How long will it take a bullet fired from a cliff at an initial velocity of 700m/s, at an

angle 30∘ below the horizontal, to reach the ground 200m below?

13. A diver in Hawaii is jumping off a cliff 45m high, but she notices that there is an

outcropping of rocks 7m out at the base. So, she must clear a horizontal distance

of 7mduring the dive in order to survive. Assuming the diver jumps horizontally, what is

his/her minimum push-off speed?

14. If Monte Ellis can jump 1.0m high on Earth, how high can he jump on the moon

assuming same initial velocity that he had on Earth (where gravity is 1/6 that of Earth’s

gravity)?15. James Bond is trying to jump from a helicopter into a speeding Corvette to capture the

bad guy. The car is going 30.0m/s and the helicopter is flying completely horizontally

at 100m/s. The helicopter is 120m above the car and 440m behind the car. How long must

James Bond wait to jump in order to safely make it into the car?

16. A field goal kicker lines up to kick a 44 yard (40m) field goal. He kicks it with an initial

velocity of 22m/s at an angle of 55∘. The field goal posts are 3 meters high.

a. Does he make the field goal?b. What is the ball’s velocity and direction of motion just as it reaches the field goal

post (i.e., after it has traveled 40m in the horizontal direction)?

17. In a football game a punter kicks the ball a horizontal distance of 43 yards (39m). On

TV, they track the hang time, which reads 3.9 seconds. From this information, calculate the

angle and speed at which the ball was kicked. (Note for non-football watchers: the projectile

starts and lands at the same height. It goes 43 yards horizontally in a time of 3.9 seconds)

Answers to Selected Problems

1. 32m

2. a. 0.5s b. 0.8m/s

3. 104m

4. t=0.60s,1.8m below target

5. 28m.

6. a. 3.5s. b. 35m;15m

7. 40m;8.5m

8. 1.3 seconds, 7.1 meters

9. 50m;v0y=30m/s;500; on the way up

10. 4.4s

11. 19∘

12. 0.5s

13. 2.3m/s

14. 6m

15. 1.4 seconds

16. a. yes b. 14m/s @ 23 degrees from horizontal

17. 22m/s @ 62 degrees

MomentumWe use the term momentum in various ways in everyday language, and most of these ways are consistent with its precise scientific definition. We speak of sports teams or politicians gaining and maintaining the momentum to win. We also recognize that momentum has something to do with collisions. For example, looking at the rugby players in the photograph colliding and falling to the ground, we expect their momenta to have great effects in the resulting collisions. Generally, momentum implies a tendency to continue on course—to move in the same direction—and is associated with great mass and speed.

Momentum, like energy, is important because it is conserved. Only a few physical quantities are conserved in nature, and studying them yields fundamental insight into how nature works, as we shall see in our study of momentum.

8.1 Linear Momentum and Force Linear MomentumThe scientific definition of linear momentum is consistent with most people’s intuitive understanding of momentum: a large, fast-moving object has greater momentum than a smaller, slower object. Linear momentum is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum is expressed as p = mv. (8.1) Momentum is directly proportional to the object’s mass and also its velocity. Thus the greater an object’s mass or the greater its velocity, the greater its momentum. Momentum p is a vector having the same direction as the velocity v . The SI unit for momentum is kg · m/s .

Momentum and Newton’s Second Law The importance of momentum, unlike the importance of energy, was recognized early in the development of classical physics. Momentum was deemed so important that it was called the “quantity of motion.” Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. Using symbols, this law is (8.7) Fnet = Δp Δt , where Fnet is the net external force, Δp is the change in momentum, and Δt is the change in time

The effect of a force on an object depends on how long it acts, as well as how great the force is. In Example 8.1, a very large force acting for a short time had a great effect on the momentum of the tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time. For example, if the ball were thrown upward, the gravitational force (which is much smaller

than the tennis racquet’s force) would eventually reverse the momentum of the ball. Quantitatively, the effect we are talking about is the change in momentum Δp. By rearranging the equation Fnet = Δp Δt to be Δp = F (8.17) netΔt, we can see how the change in momentum equals the average net external force multiplied by the time this force acts. The quantity Fnet Δt is given the name impulse. Impulse is the same as the change in momentum.

We start with the elastic collision of two objects moving along the same line—a one-dimensional problem. An elastic collision is one that also conserves internal kinetic energy. Internal kinetic energy is the sum of the kinetic energies of the objects in the system. Truly elastic collisions can only be achieved with subatomic particles, such as electrons striking nuclei. Macroscopic collisions can be very nearly, but not quite, elastic—some kinetic energy is always converted into other forms of energy such as heat transfer due to friction and sound. One macroscopic collision that is nearly elastic is that of two steel blocks on ice. Another nearly elastic collision is that between two carts with spring bumpers on an air track. Icy surfaces and air tracks are nearly frictionless,more readily allowing nearly elastic collisions on them.

Elastic Collision An elastic collision is one that conserves internal kinetic energy.

Internal Kinetic Energy Internal kinetic energy is the sum of the kinetic energies of the objects in the system.

8.5 Inelastic Collisions in One Dimension

We have seen that in an elastic collision, internal kinetic energy is conserved. An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). This lack of conservation means that the forces between colliding objects may remove or add internal kinetic energy. Work done by internal forces may change the forms of energy within a system. For inelastic collisions, such as when colliding objects stick together, this internal work may transform some internal kinetic energy into heat transfer. Or it may convert stored energy into internal kinetic energy, such as when exploding bolts separate a satellite from its launch vehicle. Inelastic Collision An inelastic collision is one in which the internal kinetic energy changes (it is not conserved).

Two objects that have equal masses head toward one another at equal speeds and then stick together. Their total internal kinetic energy is initially mv2 (1/ 2 mv2 + 1 /2 mv2) . The two objects come to rest after sticking together, conserving momentum. But the internal kinetic energy is zero after the collision. A collision in which the objects stick together is sometimes called a perfectly inelastic collision because it reduces internal kinetic energy more than does any other type of inelastic collision. In fact, such a collision reduces internal kinetic energy to the minimum it can have while still conserving momentum. Perfectly Inelastic Collision A collision in which the objects stick together is sometimes called “perfectlyinelastic.”

Basic Momentum Problems (round all final answers to nearest tenth)

1. Calculate the momentum of a 12ookg car with a velocity of 25m/s.

p = mv = 1200 X 25 = 30,000kg.m/s

2. What is the momentum of a child and wagon if the total mass of the

child and wagon is 22kg and the velocity is 1.5m/s?

p = mv = 22 X 1.5 = 33kg.m/s

3. The parking brake on a 1200kg automobile has broken, and the vehicle has reached

a momentum of 7800kg.m/s. What is the velocity of the vehicle?

V = p/m = 7800/1200 = 6.5m/s

4. A toy dart gun generates a dart with .140kg.m/s momentum and a velocity

of 4m/s. What is the mass of the dart in grams? (hint: figure kg, then convert answer

to grams)

M = p/v = .140/4 = .035kg conversion: .035 X 1000 = 35grams

5. A bowling ball of 35.2kg, generates 218 kg.m/s units of momentum. What

is the velocity of the bowling ball?

V = p/m = 218/35.2 = 6.2m/s

6. A school bus traveling at 40 km/hr. (11.1m/s) has a momentum of 152625 kg.m/s.

What is the mass of the bus?

M = p/v = 152625/11.1 = 13,750kg

Conservation of Momentum Problems (Collision Problems)

7. A 12,000kg. railroad car is traveling at 2m/s when it strikes another 10,000kg.railroad

car that is at rest. If the cars lock together, what is the final speed of the two

railroad cars?

p1 = p2

m1 v1 = m2 v2

(12,000) (2) = (22,000) v2 m2 =mass of both cars 12,000 + 10,000

24,000 = 22,000 v2

24,000/22,000 = v2

v2 = 1.1m/s

8. A 9,300 kg. railroad car traveling at a velocity of 15m/s strikes a second boxcar

at rest. If the two cars stick together and move off with a velocity of 6m/s, what is the mass

of the second car?

p1 = p2

m1 v1 = m2 v2

(9,300) (15) = (m2) (6) m2 = mass of both cars = 9,300 + X

139,500 = (9,300 + X) (6) X = mass of second boxcar

139,500 = 55,800 +6X

139,500 - 55,800 = 6X

83,700 = 6X

83,700/6 = X

X = 13,950 kg

9. A 25 gram bullet is fired from a gun with a speed of 230m/s. If the gun has a mass

of .9kg. what is the recoil speed of the gun?

p1 = p2

m1 v1 = m2 v2 Uses Newton's Third Law (action = reaction)

(.025) (230) = (.9) v2 Convert 25grams to kg = 25/1000 = .025kg

5.75 = (.9) v2

5.75/.9 = v2

6.4m/s = v2

10. A 20 gram bullet traveling at 250m/s strikes a block of wood that weighs 2kg.

With what velocity will the block and bullet move after the collision?

p1 = p2

m1 v1 = m2 v2

(.020) (250) = (2.02) v2

5 = (2.02) v2

5/2.02 = v2

2.5m/s = v2


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