Notes/Solutions on the PHYS 101 Midterm for questions that people commonly missed.
Not in all cases I just use g = 10.0 instead of 9.8 since it makes calculations easier.
Question 1:
In general this question was not done very well with a wide range of performance and
many made the question more complicated than it was and also demonstrated a false
assumption (detailed below). The Key to solving this problem is to work it out the three
relevant but separate time intervals: 0-6 seconds; 6-8 seconds; 8-20 seconds.
A force is applied to an object with initial position and velocity = 0. At time t = 6
seconds the object has a velocity of 24 cm/sec. At time t = 8 seconds the object has a
velocity of 8 cm/sec. At time t = 20 seconds the object has a velocity of 8 cm/sec.
a) In which period of time has the greatest force been applied?
 well force is manifest by acceleration so which time period has the greatest rate of
velocity change?
From 0-6 seconds the velocity went from 0 to 24 or an average acceleration of +4
(24/6) cm/sec^2
From 6-8 seconds the velocity went from 24 to 8 or an average acceleration of -8 (16/2) cm/sec^2
There is no acceleration after 8 seconds
b) What is the average velocity of the object over the period of 20 seconds?
There are two ways to do this: The hard way is shown below – the easy way comes
after you do part c)
Average velocity can be defined as: (initial velocity + final velocity)/time period
So in first 6 seconds the average velocity is (0+24)/6 = 4 cm/second
From 6 to 8 seconds the average velocity is (24+8)/2 = 16 cm/second
From 8 to 20 seconds the average velocity is 8 cm/sec
So there are three average velocities that have to be averaged in some weighted manner
over the 20 seconds. I didn’t expect anyone to do this – I did expect people to realize
the concept of average velocity might be complicated in this example
c) What is the position of the object at t = 8 seconds and at t = 20 seconds?
Now we define average velocity as total displacement/total time we then have to
compute the position of the object at 20 seconds. First what is the position at 8
seconds.
Need to use the position equation from lab 2.
So from 0-6 seconds the position is: 0 + 0*6 + ½(4)(6)^2 =72
From 6-8 seconds the position is: 72+2*24+1/2(-8)(2)^2 = 72+48-16 = 104 (the 24 is
the initial velocity at 6 seconds and while negative acceleration is applied the velocity
remains positive at all times so the object continues to move forward in the +x
direction. Some thought that the negative acceleration meant the position also changes
sign – does your car move backwards when you apply the brakes?)
So the total displacement after 8 seconds is 104 (or an average velocity of 104/8 = 13)
From 8 to 20 seconds (a=0) so the position is simply 104 + 8*12 = 200
So the total average velocity over 20 seconds is 200/20 = 10
This question was done fairly well: Blue vector = friction force (which must be
overcome for the cow to slide); Red Vector = net applied force; Yellow vector = force of
gravity. Since there are two forces at work here, the net acceleration changes so the
acceleration is not constant as would be the case if there was no friction. This produces
the curvature in the position diagrams.
3. A car of mass 1000 kg is moving at 20 m/s. The driver steps lightly on the break
for 10 s and slows the car with constant deceleration down to 10 m/s.
a) change in momentum is momentum before brakes – momentum after breaks
so that is 20*1000 – 10*1000 or 10,000
b) force is change in momentum/time = -10000/10 seconds or -1000 N
4. Uniform circular motion = the velocity is constantly changing direction therefore
a force must be at work
5. Mass is a measure of resistance to force
6. You dropped a 5 kg bowling ball from a distance of one meter on your Aunt
Harriet's precious tile floor and it didn't bounce at all:
a) how much energy was absorbed by the floor?
b) in physical terms, what aspect of the floor prevented any bounce back?
c) in what ways can you imagine that the impact of the bowling ball energy
was dissipated?
This is really a more straightforward question than many of you though. Not much
imagination was applied to part C in general.
a) all of it (50 Joules if you wanted to do the calculation)
b) Floor in inelastic – absorbs all the energy
c) 100% of the energy is dissipated as heat, sound and broken tiles
What is the principle form of energy loss in most physical systems? HEAT
8. Referring again to the cow on roller skates sliding down the incline as in problem
2. We can do an experiment and decrease the mass of the cow to find that the time it
takes to slide down the incline is the same - that is, this time does not depend upon
the mass. In terms of the vectors shown in problem 2, explain why the time it takes
to slide down and incline is independent of the mass of the object that is sliding.
The applied force depends on gravity. The frictional force depends on the normal
force. Both scale exactly the same way with Mass so mass cancels out.
13. Particle one has a mass of 2 kg and velocity of 10 m/s. Particle two has a mass of
4 kg and a velocity of -5 m/s. The two particles collide and stick. What is the velocity
of the stuck particles? In Joules, how much kinetic energy was lost during the
collision?
Momentum of particle 1 = 2*10 = 20
Kinetic Energy of particle 1 = ½(2)(10)^2 = 100 Joules
Momentum of particle 2 = 4*-5 = -20
Kinetic Energy of particle 2 = ½(4)(5)^2 = 50
Total momentum of system = 0
Total energy of system = 150 Joules
Upon collision the total momentum for the stuck particles has to be zero so the velocity is
zero so all (150 Joules) of kinetic energy are lost.
14. A train moving on a frictionless track has a mass of 10,000 kilograms. The train
is moving at 1 meter per second. Another train of mass 20,000 kilograms is sitting
on the track and the first train collides with it so they stick together. What is the
approximate velocity of the two trains, now stuck together?
Again apply conservation of momentum – many people just didn’t do this.
Momentum before = mv = 10,000*1
Therefore the momentum after has to be 10,000
Combined mass =30000;
mv = 10000; v = 10000/30000 = 1/3
16. A car moves along a frictionless track with a horizontal acceleration of 8
centimeters per second -2.
The cart has an initial velocity of 20 cm/s and has an initial position of xo = 2cm.
a) What is the velocity of the cart at time = 5 seconds?
b) What will be the total distance traveled at time = 5 seconds?
Generally the second part of this was done poorly.
a) v = v_o +at; = 20 +8*5 = 60 cm/sec
b) p = p_o +v_ot + 1/2at^2 = 2 +20*5 +1/2(8)(5)^2) = 2+100+100 = 202
202 = the position but we started at 2 so the total distance traveled would be 200 cm
Questions 17 through 19 were not well at all; a few of you said they couldn’t be solved.
They are all solved the same way – conservation of energy. 1/2mv^2 = mgh so v^2 =
2gh is all you need
17. Compute the impact velocity of an object that is dropped from a height of 12
meters above the surface. (show your work)
V^2 = 2gh; v^2 = 2*10*12 v = sqrt(240) (about 15.5 m/s)
18. I toss a ball into the air with an initial velocity of 10 m/sec. What is the
average velocity of the ball during the time it rises to its maximum height.
Well at its maximum height the velocity will be 0
So average velocity is (10+0)/2 = 5 m/sec
19. I throw a ball into the air with an initial velocity of 3 m/s. How high will the
ball rise? How long will it take to return to my hand?
a) v^2 = 2gh; h = v^2/2g = 9/20 = 0.45 m
b) how long does it take to drop in free fall from 0.45 m ?
d = 1/2at^2; t^2 = 2*.45/10 = sqrt(.9) or about 0.3 seconds. Total return to hand
time is then 2x0.3 = 0.6 seconds
A skier of mass 70 kg starts from rest at the top of a hill with vertical height = 20
meters. The skier reaches the bottom of the hill with a speed of 10 m/s. How much
energy was lost by the skier, due to friction, in their journey from the top to the
bottom of the hill?
PE (top of hill) = mgh = 70*10*20 = 14000 Joules
KE at bottom = 1/2mv^2 = ½(70)(10)^2 = 3500 Joule
14000- 3500 = energy loss.