Download Semester Exam REVIEW PACKET KEY

Physics Review: Semester 1 Circle the description to the right that best describes the graph on the left.
NOTE: If you choose ‘Not Moving’ you can skip ahead to the next graph.
Decide if you are:
5.0
Moving
Not moving
4.0
D
3.0
If you chose ‘Moving’, then select the direction:
2.0
Towards
1.0
0.0
0
2
4
6
t
8
10
Away
or
Speeding up
Describe your speed:
Slowing down
Moving at a constant speed
Decide if you are:
Moving
5.0
Not moving
4.0
D
If you chose ‘Moving’, then select the direction:
3.0
Towards
2.0
Away
or
1.0
0.0
0
2
4
6
t
8
10
Speeding up
Describe your speed:
Slowing down
Moving at a constant speed
Decide if you are:
Moving
5.0
Not moving
4.0
D
3.0
If you chose ‘Moving’, then select the direction:
2.0
Towards
or
Away
1.0
0.0
0
2
4
6
t
8
10
Describe your speed:
Speeding up
Slowing down
Moving at a constant speed
Decide if you are:
Moving
5.0
Not moving
4.0
D
If you chose ‘Moving’, then select the direction:
3.0
Towards
2.0
Away
or
1.0
0.0
0
Speeding up
Describe your speed:
2
4
6
t
8
10
Slowing down
Moving at a constant speed
Decide if you are:
Moving
5.0
Not moving
4.0
D
If you chose ‘Moving’, then select the direction:
3.0
Towards
2.0
Away
or
1.0
0.0
0
Speeding up
Describe your speed:
2
4
6
t
8
10
Slowing down
Moving at a constant speed
Decide if you are:
Moving
5.0
Not moving
4.0
D
If you chose ‘Moving’, then select the direction:
3.0
Towards
2.0
or
Away
1.0
0.0
0
2
4
6
t
8
10
Describe your speed:
Speeding up
Slowing down
Moving at a constant speed
Decide if you are:
Moving
5.0
Not moving
4.0
D
If you chose ‘Moving’, then select the direction:
3.0
Towards
2.0
Away
or
1.0
0.0
0
2
4
6
t
8
Speeding up
Describe your speed:
10
Slowing down
Moving at a constant speed
Decide if you are:
Moving
Not moving
4.0
2.0
V
0.0
If you chose ‘Moving’, then select the direction:
2
4
6
8
10
t
Towards
Away
or
‐2.0
Speeding up
Describe your speed:
‐4.0
Slowing down
Moving at a constant speed
Decide if you are:
Moving
4.0
Not moving
2.0
V
0.0
2
4
6
8
10
t
If you chose ‘Moving’, then select the direction:
Towards
Away
or
‐2.0
Speeding up
Describe your speed:
‐4.0
Slowing down
Moving at a constant speed
Decide if you are:
Moving
4.0
2.0
V
0.0
2
4
6
8
10
t
Not moving
If you chose ‘Moving’, then select the direction:
Towards
or
Away
‐2.0
‐4.0
Describe your speed:
Speeding up
Slowing down
Moving at a constant speed
Decide if you are:
Moving
4.0
2.0
V
0.0
2
4
6
8
10
t
Not moving
If you chose ‘Moving’, then select the direction:
Towards
Away
or
‐2.0
Speeding up
Describe your speed:
‐4.0
Slowing down
Moving at a constant speed
Decide if you are:
Moving
Not moving
4.0
2.0
V
0.0
If you chose ‘Moving’, then select the direction:
2
4
6
8
10
t
Towards
Away
or
‐2.0
Speeding up
Describe your speed:
‐4.0
Slowing down
Moving at a constant speed
Decide if you are:
Moving
4.0
Not moving
2.0
V
0.0
2
4
6
8
10
t
If you chose ‘Moving’, then select the direction:
Towards
Away
or
‐2.0
‐4.0
Speeding up
Describe your speed:
Slowing down
Moving at a constant speed
Decide if you are:
Moving
4.0
2.0
V
0.0
2
4
6
8
10
t
Not moving
If you chose ‘Moving’, then select the direction:
Towards
or
Away
‐2.0
‐4.0
Describe your speed:
Speeding up
Slowing down
Moving at a constant speed
I.
Graphing a. What does the slope of a position vs. time graph represent? Velocity b. What does the slope of a velocity vs. time graph represent? Acceleration c. What does the area under the curve of a velocity vs. time graph represent? Displacement d. From the given graph in each of the sets below, sketch the shape of the corresponding graphs. tim
tim
position
tim
tim
tim
velocity
velocity
tim
tim
acceleration
velocity
iii.
position
position
ii.
acceleration
i.
acceleration
tim
tim
8
6
4
2
v
(m/s) 0
-2
-4
2
4
6
t
(s)
-6
-8
e. From the graph above… i. What is the change in velocity from t = 0 s to t = 6 s? Δv = vf – vi = 4m/s – (‐4 m/s) = 8 m/s ii. Determine the displacement of the object in the first 2 seconds d = vit = (‐4m/s)(2 s) = ‐8 m iii. Determine the displacement of the object from t = 2 s to t = 6 s d = vit +1/2at2 and a = Δv/Δt = 8 m/s / 4s = 2 m/s2 d = (‐4)(4) + ½(2)(42) = 0 m iv. Determine the objects acceleration from t = 2 s to t = 8 s As shown above in part iii, it is still 2 m/s2 II. Motion a. Jenn walks from her home East 20 m, then realizes that she forgot her wallet in her room. She then runs back towards her house West 15m. As she is running, she finds her wallet in her back pocket! She then continues heading east for another 25 m. i. What is the distance that Jen travelled? d = 20 m + 15 m + 25 m = 50 m ii. What is the displacement that Jen travelled? d = 20 m – 15 m + 25 m = 30 m East iii. Why are the answers to (i) and (ii) different? Displacement is a vector and direction must be included b. Define the term speed AND in addition, give a mathematical equation to describe speed. Speed is the rate of distance travelled, or how fast an object is travelling s = Δd/Δt c. Define the term velocity AND in addition, give a mathematical equation to describe velocity. Velocity is the rate of the displacement travelled, and can also be described as how fast you’re travelling in a specific direction. v = Δd/Δt d. An object travels 10m in one second, then 10m in the next second, then 10m in the next second i. What is the object’s velocity? v = 10 m/s ii. What is the object’s acceleration? a = 0 m/s2 e. Define the term acceleration AND in addition, give a mathematical equation to describe acceleration. Acceleration is the rate of velocity change a = Δv/Δt f.
Graphing: Draw a displacement vs. time, velocity vs. time, and acceleration vs. time graph for the following scenarios i. An object at rest ii. John walking at a constant velocity in the negative direction iii. A car, starting at position x = 0, accelerating from rest in the positive direction iv. A small plane that always has a positive velocity and a negative acceleration v. A biker always having a negative velocity with a constant positive acceleration g. Using the graph below…. A
x
(m)
B
0
5
t(s)
i. Which object is travelling faster at t = 5 s? How do you know? A, because the slope is the velocity and A has a steeper slope. It must therefore be moving faster ii. What is happening at t = 5 s? The two objects are at the same displacement or position iii. At t = 0, who is faster? Who has a greater positive starting position? Still A (see explanation in part i) and B has the greater positive starting position III. Projectiles a. What is a projectile? Airborne object only affected by the force of gravity b. A plane is trying to travel north at a velocity of 405 km/hr, but experiences a 75 km/hr crosswind blowing from east to west i. Draw a diagram of the plane’s motion ii. When taking into consideration the crosswind, what resultant velocity (calculate speed and calculate direction!) will the plane be travelling at? R = √(752 + 4052) = 412 km/hr iii. After two hours of this weather, how far north will the plane have travelled? d = vt = (405 km/hr)(2 hr) = 810 km iv. How long will it take the plane to travel 405 km west? t = d/v = 405 km/75 km/hr = 5.4 hr c. A ball is thrown straight up into the air on the planet earth i. At the top of the ball’s motion, what is the ball’s velocity? 0 m/s ii. At the top of the ball’s motion, what is the ball’s acceleration? ‐10 m/s2 iii. During the time the ball is in the air, what never changes, acceleration or velocity? Why? Acceleration, because the surface gravity on Earth stays the consistent d. A rocket is shot off of a 40.0 m platform horizontally with a velocity of 250. m/s. i. How long is the rocket in the air for? d = 1/2gt2 or t = √(2d/g) t = √[(2)(40m)/(10 m/s2)] = 2 s ii. How far away from the base of the platform does the rocket travel? d = vt = (250 m/s)(2 s) = 500 m e. A soccer player kicks a ball that travels at an angle of 35.0 degrees to the horizontal with a velocity of 12.5 m/s i. What is the ball’s vertical velocity component? vy = vsinθ = (12.5 m/s)(sin 35) = 7.17 m/s ii. How long is the ball in the air for? vf = vi + atup or tup = (vf‐vi)/a tup = (0 m/s – 7.17 m/s)/‐10 m/s2 = 0.717 s ttotal = 0.717 s x 2 = 1.43 s iii. How high does the ball go at its highest point? d = vit + 1/2at2 d = (7.17 m/s)(0.717 s) + 1/2 (‐10 m/s2)(0.717)2 = 2.57 m iv. What is the ball’s horizontal velocity component? vx = vcosθ = (12.5 m/s)(cos 35) = 10.2 m/s v. How far away from the soccer player does the ball land? d = vt = (10.2 m/s)(1.43 s) = 14.6 m IV. Forces a. Describe the following types of forces i. Gravitational Force due to gravity (weight) ii. Tension Force due to a pull from a string/chain/rope etc. iii. Normal Contact force perpendicular to surface iv. Friction Force that opposes the motion of two objects rubbing together b. What are the units for forces? Newtons c. What two pieces of information can we gather from a force vector? Magnitude and Direction d. What is the smallest vector that can be formed when adding force vectors of 4N and 7N? What is the largest? Smallest – 3N, Largest – 11N e. Write out Newton’s 3 Laws in YOUR OWN WORDS and give two examples of each, explicitly explaining how each example ties in to Newton’s laws. 1st Law: Inertia (an object in motion will stay in motion and an object at rest will stay at rest until an outside force acts upon it) 1. plates and cups stay on a table while a tablecloth is quickly removed under it 2. A driver without a seatbelt flies through the windshield when the car comes to a sudden halt after a crash 2nd Law: F = ma A car with a large mass will have a small acceleration compared to a car with a small mass with the same force (same engine!) 3rd Law: forces come in pairs that are equal in magnitude and opposite in direction When a small car and a large truck collide, they will both experience the same force. f.
What is the difference between mass and weight? Give a mathematical equation that represents what weight is. Mass is the amount of matter in an object, weight is the force of gravity acting on an object. Weight = Fg = mg g. A small car collides head‐on with a large truck. i. Which vehicle will undergo a greater acceleration? Why? The small car because it has a smaller mass ii. Which vehicle will experience more force? Why? Both will experience the same force because they are a 3rd Law action‐reaction pair with no outside forces acting on the cars h. A small car and a large truck are given the same engine (which generates the same amount of force!) Using Newton’s second law, explain why the large truck should have a slower acceleration. Both cars have the same force, and since the large truck has a greater mass, the acceleration must be less i.
Two small cars are given two different engines (which generate different amounts of force!) Using Newton’s second law, explain which car will accelerate faster The car with the more powerful engine will accelerate faster since their masses are the same and the force will be greater from the larger engine j.
A 15kg crate is pulled along frictionless ice with a force of 45N. i. Draw a complete force diagram for the crate ii. What is the crate’s acceleration? a = F/m = 45 N/ 15 kg = 3 m/s2 k. A 4.00kg block is dragged along rough carpet with a tension force of 52N and a force of friction of 8.00N. i. Draw a complete force diagram for the block ii. What is the block’s acceleration? a = F/m = 44 N/ 4 kg = 11 m/s2 l.
A 35kg girl swings on a seat attached to three vertical ropes. i. Draw a complete force diagram for the girl on the seat ii. How much tension is in each of the three strings? 117 N m. Is it possible to move a book across a table without a net force? How? Yes! If the push on the book is equivalent to the force of friction on the book, the book can move with a constant velocity V. Momentum a. What is momentum? Moving Inertia: p = mv b. If an object had a mass of 14.0 kg and a momentum of 1250 kgm/s, find the velocity of the object v = p/m = 1250 kgm/s/ 14 kg = 89.3 m/s c. A cannonball is shot from a cannon. Describe the relative motions of each object after the cannonball is shot out of the cannon and why they move in such a manner. The cannonball moves in one direction and to conserve momentum (since before the explosion, the overall momentum was 0!) the cannon moves in the opposite direction d. What is impulse? Impulse is the change in momentum that happens to an object. I = Ft = mΔv e. If a 22.5kg medicine ball that was moving at 5 m/s is slammed to the ground with a force of ‐40,500 N, find the time of impact on the ball. t = Δp / F = [0 kgm/s – (22.5 kg)(5 m/s)] / ‐40,500 N = 0.00278 s f.
A wide receiver, whose mass is 85.0 kg, is running at 11.0 m/s when he catches a football thrown in his direction traveling at a velocity of ‐27.5 m/s. If the football has a mass of .950 kg, find the velocity of the wide receiver and football after the catch. Inelastic Collision : m1v1 + m2v2 = (m1 + m2)v1+2 (85)(11) + (0.95)(‐27.5) = (85 + 0.95)v1+2 v1+2 = 10.6 m/s g. Car A (mass = 1250kg), traveling at 10.5 m/s, hits a stationary Car B (mass = 2050kg) and bounces off. If Car B moves forward at a velocity of 2.1 m/s, find the velocity of Car A after the collision. Elastic Collision: m1v1 + m2v2 = m1v1’ + m2v2’ (1250)(10.5) + (2050)(0) = (1250)v1’ + (2050)(2.1) v1’ = 7.06 m/s VI. Gravity a. An asteroid with mass M1 is a distance R away from earth, whose mass is M2 i. If the asteroid were to travel farther away from earth, what would happen to the force of gravity between the two objects? The force would decrease between the two objects ii. If M1 quadrupled in quantity, what would happen to the force of gravity between the two objects? Force of gravity would be 4 times as much iii. If R tripled, what would happen to the force of gravity between the two objects? Force of gravity would be 1/9th of its original force iv. If M1 doubled, M2 doubled, and R doubled, what would happen to the force of gravity between the two objects? Force of gravity would remain the same!