Physics 218 Midterm 2Practice ExamDr. Edward Rhoads
______Name
Multiple Choice (2 points each):
Please circle your answers.
Good luck!
1)A wheel with a radius of 5 m and mass 0.2 kg is given an angular acceleration of 2 rad/s2. After 4 seconds what will the angular velocity of the wheel be?
a)0.8radians/sec
b)8 radians/sec
c)40 radians/sec
d)20 radians/sec
2)A wheel with a radius of 5 m and mass 0.2 kg is given an angular acceleration of 2 rad/s2. After 4 seconds how far (in terms of angle) will the wheel have rotated?
a)1.6 radians
b)16 radians
c)160 radians
d)80 radians
3)A 70 kg person is initially at rest and standing on a 14 kg rug which is on top of a frictionless floor. If that person then walks forward and walks off the rug with a forward velocity of 0.2 m/s then what velocity will the rug have?
a)0.2 m/s forwards
b)1 m/s backwards
c)2 m/s backwards
d)0.2 m/s backwards
4)A green car of 1000 kg runs into a yellow car of equal mass which is initially at rest. If the green car was traveling 8 m/s forward before the collision and the yellow car moves 10 m/s forward after the collision then the motion of the green car is:
a)backwards
b)no motion
c)forwards
d)it is impossible to tell from the information given
5)A ball is thrown northwards and bounces off of a wall. If the ball was moving at 5 m/s North before it hit the wall and moves 5 m/s South after it hits the wall then the change in momentum is:
a)Positive in the South direction
b)Zero
c)Positive in the North direction
d)impossible to determine based on the information given
6)A light bulb with a power of 12 Watts is turned on for 5 seconds. How much energy does that light bulb consume?
a)12 J
b)60 J
c)300 J
d)5 J
7)A 5 kg ball is launched from the ground with a kinetic energy of 200 J. When the ball reaches its highest height then how much potential energy will the ball have?
a)0 J
b)200 J
c)400 J
d)1000 J
8)A box is pushed along a frictionless surface with a force of 12 N for a distance of 15 m. The force is then removed. If the mass of the box is 10 kg and its initial kinetic energy is 20 J then what will the kinetic energy of the box be when the force is finished?
a)2000 J
b)180 J
c)200 J
d)20 J
9)A weight at the end of a horizontal rod of negligible mass produces a torque of 2000 Nm clockwise. A string is tied to the center of the rod. What is the torque caused by the tension of the rod if the length of the rod is 4 m. Assume the rod does not rotate.
a)2000 Nm counterclockwise
b)500 Nm counterclockwise
c)1000 Nm counterclockwise
d)The tension will produce no torque
10) A ball has an initial kinetic energy of 200 J and initial potential energy of 300 J. 400 J of work is then done on the ball. What will the total energy of the ball be after the work is done to it?
a)900 J
b)500 J
c)400 J
d)It is impossible to determine without knowing the mass of the ball
SHORT ANSWER: (20 points each)
Be sure to answer every part of every question. Be sure to label the parts of questions you are answering. Be sure to show work and include units and directions. The lowest of the 5 short answers will be dropped.
1)Head on collision. A truck traveling north crosses a meridian and runs head on into a car moving south. Before the collision the 2000 kg truck was traveling 18 m/s North and the 700 kg car was traveling 32 m/s South.
A) Initially what is the combined (total) momentum of both vehicles together?
B) After the collision the truck moves North at 8 m/s. What will the velocity of the car be after the direction (hint, what will the total momentum after the collision be)?
C) What is the impulse (aka change in momentum) that the car receives during the collision?
D) Using the impulse found in C what is the Force acted on the car if the impact time is 0.7 seconds?
2)The Stuntman rocket cycle jump.
A stuntman has a rocket cycle that blasts out hot air in order to push him forward. In the middle of a frictionless lake he attempts a stunt. He throttles his engines to full as the 400 kg cycle (which includes the mass of the stuntman on it) produces a force of 5000 N.
A) The cycle starts at a distance of 60 m from a launch ramp. What is the work that the engine does on the cycle in that distance?
B) When the cycle reaches the ramp how fast is it going (i.e. what will its velocity be, and you don’t need a direction here)?
C) The cycle will launch from the ramp going straight up at the velocity it entered the ramp. The cycle leaves the ramp at a height of 24 m above the ground. What is the maximum height above the ground that the cycle will reach?
D) Before the race the stuntman gives his cycle one final test. He finds that at full throttle that the engine will produce a work of 150,000 J of power in 32.2 seconds. What is the power of his engine?
3)Chain reaction
A 700 kg car is parked at the top of a hill. However, its driver forgot to activate the parking break! So, the car will roll down the roll. The vertical height the car rolls down is 23 m.
A) What will the velocity of the 700 kg car be when it reaches the bottom of the hill if you ignore friction?
B) At the bottom of the hill the 700 kg car runs into a 500 kg car. The two cars stick together. What will the velocities of the two cars be after the collision? Again ignore friction.
C) Will this collision be an elastic collision or will it be an inelastic collision? Explain why.
D) After the collision friction does -3000 J of work on the two cars (total not each). The cars will then roll up a hill. How far up (vertically speaking) the hill will the two cars roll (assume they stay stuck together)?
4)Around the bend.
An old freeway has a flat turn that has an effective radius of 220 m. The road, being old, has a frictional coefficient of 0.22. A 800 kg car enters the turn.
A) If the car enters the turn at 24 m/s (which is about 54 mph) then what will the centrifugal force on the car be?
B) The sign warns motorists to go 45 mph around the turn (which is 20 m/s). Is this a save velocity to go on this turn? Why or why not?
C) The next day it rains. With the oil on the road and the water the road becomes a bit slippery and the frictional coefficient is dropped to 0.12. For this condition what is the fastest that a motorist would want to drive on the road?
D) A driver on this rainy day does not drive based on the weather conditions. In fact they try to speed through the turn at 25 m/s. The car has a mass of 700 kg. The car skids going into the corner and slams into an experimental spring powered guard rail at 25 m/s. The spring coefficient of the guard rail is 7500 N/m. What is the distance it will take for the guard rail to stop the car (ignore friction)?
5) Tortoisevs. the Hare the wheel race.
The next race involves a giant wheel. Each will apply a force to create a torque to spin the wheel. The setup:
The Hare: radius = 0.8 m. Moment of Inertia (I) = 0.45 kg m2 Starts at an angular velocity of 0.78 rad/sec. Maintains constant angular velocity by applying a torque to counter out air drag and friction.
The Tortoise: radius = 1.2 m. Moment of Inertia (I) = 0.56 kg m2 Starts at rest.
A) The tortoise applies a 15 N force on the wheel at the very outside of the wheel. Find the Torque then use that torque to find the angular acceleration of the wheel.
B) What is the linear velocity of the Hare’s wheel?
C) If the Tortoise finishes the race in 16 seconds then how much has the wheel rotated in radians?
D) What will the final angular velocity of the Tortoise be (please do not give the average angular velocity)?
E) Who will win the race?
EQUATIONS:
X = Xo + Vo t + ½ at2
V = Vo + at
W = mg
F = ma
F = μN
F = G M1 M2 / r2 (towards object creating force)
G = 6.67 * 10-11 N m2 / kg2
g = 9.80 m/s2
Net Force = sum of all forces
Up = - Down
sin(theta) = opposite / hypotenuse
cos(theta) = adjacent / hypotenuse
tan(theta) = sin(theta) / cos(theta) = opposite / adjacent
c2 = a2 + b2
Quadratic Equation: x = [-b + (b2 – 4ac)1/2]/2a
p=mv
p total = m1v1 + m2v2+ …
total momentum before collision = total momentum after collision
Impulse = change in momentum = m * change in v = F * t
Net work = change in total energy = net force * distance
W = F d
kinetic energy = ½ mv2
PE = mgh
P = W/t = energy / time
F = -kx
U = 0.5 kx2
θ = θ0 + ω* time + ½ α t2
ω = ω0 + α t
X = θ R
v = ω R
a = α R
L = Iω = mvr
Torque = F * R = I α
F = m v2/r
a = v2/r
Orbital velocity = [GM/r]1/2
Period2 = [4 π2 /(GM)] * r3