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Solutions for the Mathematical Physics Unit.
You will derive much more long term benefit from this unit if you first try to solve the problems
without looking at these solutions, hints and answers. You need experience struggling with a
hard problem and figuring it out either on your own or with others.
Day 1:
 The difficulty with this unit will be the math – not the physics. This particular class is based on d
= vt which is easy physics and yet the problems are hard. You need to excel in this unit if you are
going into physical sciences.
 The displacement is the area under the v-t graph.
 Focus on what you are supposed to calculate and don’t calculate other stuff.
 Make estimates or limits of the answer so you can see if it is reasonable.
 Sketch graphs and diagrams. Turn these into equations. Solve for the system of equations.
 Use variables whose symbols mean something – not just x and y. Include subscripts.
1) The answer will be less than 100 and more than 60. The answer will be less than 80 because more time
is spent at the slow speed. Answer 72 km/h. If you sketch the v-t graph it might help you realize this.
Find total distance during each time interval and divide by 10. Answer: 72 km/h.
2) The answer is a ratio and it should be smaller than 1 because Millie will have a higher average because
she spends more time at the high speed. A sketch of v-t graphs might help you realize this. Use v for
the walking speed and 2v for the running speed – it will look simpler than using subscripts
everywhere. Set up the equation for the sum of Willie’s two times and for the sum of Millie’s two
distances. Solve these for the two times in a ratio of the two. Answer: 8/9
3) Make a sketch of Waterloo, Guelph and where they meet at noon which is NOT in the middle. Form
an equation for each distance using Millie’s speed and Willie’s speed. Solve the two equations for the
time from the start until they meet. Answer: t = root 2.25 = 1.5 hours before noon. .
4) Draw a diagram of the two races and set up equations. Remember that you need the ratio of the times –
not the actual values. Answer: 24/25
5) This is very similar to question 2! This time you are given the distance and have to find the actual
times – not just a ratio. Answer: The British take 28 hours and the French take just under 23 hours.
6) This is hard. Draw a picture of where Willie and the train get to each time.You are looking for a ratio.
Set this up using what you know about Willie’s speed and the two unknown times. The key to the
solution is realizing that the difference between the train’s distances each time is AB. This means you
can use the train’s information to find how the two times are related. Answer: 1/(1 + 3/2) = 2/5
Day 2:
The two videos are just there to provide an aid to visualizing the different problems coming up.
1) Set up the three d = vt equations and solve for the unknown time. Answers: 36 s, impossible
2) Draw a diagram of the displacements. Find the velocity with respect to the shore for the first boy and
then find the time to go 40m. Find the time for the other to go across, how far downstream he lands
and then how long it takes to run back. Answer: 23.1 s - (20 s + 3.3 s) = -0.3 s. However, in terms of
significant digits and a realistic situation, this would be a tie.
3) This is very similar to the first question. Answer: 7.5 h
4) In the raft frame of reference, the time there and back must be the same. This must also be true for the
shore frame. There for the raft moved 1 km in half an hour. Answer: 2 km/h
5) This is just like the previous question. Answer: 3 km/h
6) Draw a diagram of the displacements. Find the speeds of Jean and Lucy in terms of the river using Vl
= 1.5 Vj and Vj+ Vr = Vl – Vr. Then find their speeds and times going straight across. 750/1.2
sqrt(15) and 750/1.2 sqrt(35). Answer: 56 s.
Day 3
 When using the equations for constant acceleration, you need to know three pieces of
information to find a fourth. Choose the equation that gets you there in one step.
 These questions may cause you to go around and around in circles because there are so many
different things you can calculate. Focus on what you know and what you want to know.
 Draw diagrams of the displacements – so you can see how they are related.
 The questions are not ordered by ease – look for the easiest questions and do them first.
 If you have not got the answer after 10 minutes – move on.
 No formula sheets allowed for the quizzes from now on.
1) Convert from km/h to m/s by dividing by 3.6. Calculate how far he goes while braking. Note that the
final speed is not given explicitly but if you are stopped then v = 0. Answer: He travels 4 m while
braking, so can travel 196 m more and terrify the protesters.
2) Catching up means that the two displacements are equal. vt = ½ a t2. Answer: 1.7 x 104 s.
3) For both parts, and end speed is zero, so d = ½ at2. The 25 ms is insignificant. Answer: 4.0 s
4) Similar to #2 except that one has a head start so that vt + d = ½ a t2. Need to use the roots of a
quadratic. Answer: t = 43.7 s and so distance is 1.9 km.
5) A simple question with irrelevant information. d = 250 + 140*22/3.6 + 140*8/2*3.6 Answer: 1.3 km.
6) A question that is long and tedious even if you do it as efficiently as possible. You want to find
acceleration, so focus on finding that. The starting speed v0 is zero and the mid and end speeds are v1
and v2. Therefore using v22 = v02 + 2ad for the whole drop and the first half, we have a = v12 /160 and
a = v22 /320. This also means that (v1 + v2)/2. We need to find one of those two speeds. Using the
information from the second half of the drop, we know that the average speed is 80/2 = (v1 + v2)/2,
therefore v1 + v2 = 80. Using this and (v1 + v2)/2 you find that v2 = 46.9 m/s. Answer: a = 6.87 m/s2
7) This is possibly the worst question. You cannot assume that the speed at the first cone is zero. Notice
how there is only a slight increase in speed over this distance. The v-t graph is shown below. The
speed will be greater than the starting value that is around 35 m/s, but not doubled. We are given
distances and times for three intervals and might consider writing equations for average velocity.
(v0 +v1) = 2*10/.30 = 66.7 m/s
(v0 +v2) = 2*20/.59 = 67.8 m/s
(v1 +v2) = 2*10/.29= 69.0 m/s
If you subtract the first two you get
(v2 - v1) = 1.1 m/s. This is v, so a = 1.1/.29 = 3.79 m/s2.
At this point we know the distance, the acceleration and want to know the speed at the end. This means
that we either need a starting speed or the time of travel. Finding any of the speeds will be easier
d= v1t+ ½ a t2
10 = v0(0.3) + ½ 3.79 (0.3)2.
v0 = 32.8 m/s
Finally, v32 = v02 + 2ad
v32 = 1076 + 758
Answer: v3 = 43 m/s
8) This one is just tedious.
Carl’s time is simplest. He runs at an average of 11/2 m/s for 1.0 s and covers 5.5 m. He covers the
remaining 94.5 m at 11 m/s which takes 8.59 s. Total = 9.59 s.
Ben runs at an average of 11/2 m/s for 0.5 s and covers 2.75 m. He covers 87.25 m at 11 m/s which
takes 7.93 s. The last 10 m use d = v2t- ½ a t2. 10 = 11t – t2. Factor or use the quadratic roots formula
to get that this takes 1.0 s. Total = 9.43 s. Answer: Ben wins by 0.16 s.
9) This one is much easier. The time to hit ground is 8.08 s. Superman only has 3.08 s to cover the same
distance, with the same acceleration and a non-zero starting velocity. Answer: 89 m/s. However, being
suddenly stopped by Superman is just as deadly as being suddenly stopped by the ground.
Day 4
The three videos give examples of projectile motion.
 Video#1: The two components are independent. Sketch the paths and the positions at three
different times. Note: This only works if the bullet is fired horizontally and there is no drag
 Video#2: Sketch the path of the ball and plane their positions at several times. The ball must be
dropped before the plane is overhead. Suppose the height is 500 m, the plane’s speed (and the
ball’s speed) is 100 m/s. How far back horizontally do you release? t ~ 10 s and you should release
1 km before. Note: This will not work in real life because air resistance will NOT be negligible.
 Video #3: Sketch the positions of the path of the ball and monkey if they do hit. Mark several
positions at equal time intervals along the path. Consider what you would see if you fell at the
same time as the monkey. What does the monkey look like? The ball?
Write the vertical information in one place and the horizontal information in another.
1) There is an initial vertical velocity - unlike the basketball problem. The vertical information lets you
find t using the roots of a quadratic formula. t = 6.21 s. Answer: v = 6.4 m/s
2) Sketch the ball’s path and the hill. The hill constraint means that the vertical displacement over the
horizontal is tan (30) = 0.577. Using the two distance formulae you find that t = 3.27 s. Use this to find
the distance along the slope – the hypotenuse. Answer: 105 m
3) Sketch and label the problem. This is a friendly triangle, so the velocities are 36 and 27 m/s and dH =
36 t = and dv = 27t - 4.9 t2 = 10. Solve the second equation for t. Take the larger root because it is for
the point on the way back down. Answer: 36 * 5.1 = 184 m.
4) Sketch the problem. The speed when it started is the same as the speed when it hits the ground. The
horizontal component stays the same throughout and is v /sqrt(2). It is also equal to the vertical speed
at the fence. Horizontal: 8 = v t /sqrt(2) and Vertical: 10 = v t /sqrt(2) + 4.9 t2. Solve for t (0.639) and
vH (12.5). Find vv = 12.5 + 0.639*9.8 = 18.8. Use Pythagoras to find the total speed. Answer: 23 m/s
5) This one is similar to the last. Horizontal: 5 = v t /sqrt(2) and Vertical: 4 = v t /sqrt(2) - 4.9 t2. Solve for
t (0.452) and vH = vv (11.1). Time to the middle, is the time that reduces the vertical velocity to
nothing (1.13). Total distance = 2*1.12*11.1=25. Answer: 25-5 = 20 m.
6) The horizontal distance is 500t. The vertical distance is 7070 ft. The flight path, is the hypotenuse = 50
t2. Use Pythagoras and you will find that the time is 14.14 s. This means the horizontal distance is
7070 – the same as the vertical. Answer: 45o.
7) This one is not too tough. The time to fall is 2.47 s and the time to drive is 6.39s. Answer: 3.9 s.
8) Vertical: 490 = 4.9 t2 and so t = 10 s. Also tan(A) = 490/range = 490/1000 . Answer: A = 26o.
Day 5
 The tug of war illustrates a 2-D force diagram similar to what is needed for questions 1 and 8.
 The pulley question illustrates that with an ideal pulley, the force of tension is equal on both sides
of the pulley. This is needed for question 2.
 The elastic band analysis is an introduction to Hooke’s Law and questions 4 and 11.
 The ladder question introduces the idea of torque and is needed for questions 3 and 7. If an object
is in equilibrium, the torques as well as the forces must be balanced. Torque is r X F. The floor has
a normal force equal to the weight which acts at the centre of mass. The normal force from the wall
counters the torque that these produce. The floor’s is needed to balance the wall’s normal force.
Draw force diagrams!!!!
1) The tension acts along the rope in both directions. The horizontal components cancel. The vertical
components add and must each support half of the weight. Use trigonometry to find the tension.
Answer: 1.2 kN
2) Pulley C has 2F down and so must have 2F up. B has 2400 down and F, F and 2F up. Therefore 4F
= 2400. Answer: 600N
3) The corner acts as a pivot point where the torques must be balanced. Two ducks provide 4*6 kg
one way and the board provides 1*5kg the other way. The difference is provided by Huey, x*3kg.
Answer: 6.3 m from the pivot point, which is farther than the end of the board! It won’t tip.
4) When the spring is chopped into thirds, each piece becomes harder to stretch, is K is tripled and the
stretch is 1/3. The first piece supports all three masses, but the second only supports 2 and the third
only 1. The stretch is 3 inches (1/3 + 2/9 +1/9). Answer: 2 inches.
5) There is no force!
6) The force diagram is like the ladder question at the top. The total force of the floor uses Pythagoras
and the friction (equal to wall’s normal) and normal (equal to weight). Answer: 1.13 kN
7) The short length of rope has a great tension because the horizontal components must be equal.
Horizontal: T1cos30 = T2cos45. Vertical: T1sin30 + T2sin45 = 19.6. Answer: 18 N
8) A force diagram with force equal to the 2 tensions, shows that the horizontal component must be
half the hypotenuse. Answer: 120o
9) At one end there is a spring force three times as big as the other end. (Same stretch and different
k.) The upward force of 4/3 F is balanced by 6 mg down. The weight of the board acts in the
middle and that of Batman is x units from the stringer spring. Balance torques. Answer: 2.0 m
10) This is just like question 4. Answer: 2.0 m
Day 6
 The eraser on the board looks at friction and inclined planes and is needed for question 7 and 8.
 The tablecloth looks at how friction has a maximum force and therefore a maximum acceleration
before slipping occurs. This is needed for question 7.
 The Slinky relates to numbers 9 and 10.
Draw force diagrams!!!!
1) This is like a cart being pulled by a mass having vertically over the table edge. It can be viewed as a 1D problem. The net force is Jane’s weight minus friction on Tarzan and it acts on the combined
masses. The textbook does this as a system of equations. Answer: 3.0 m/s2.
2) It doesn’t matter what the local gravity is – just don’t assume that it is 9.8 N/kg. There is an upward
force of 165 N and the down ward will still be the 110 N that was balanced earlier. Answer: 2.5 m/s2.
3) There are three possible systems. The small block, large one and total. Answer: 1.0 N
4) Find the acceleration of 31.6 m/s2 = 3.22 g’s. There is one g down, so the scales are provided 4.22 g’s.
Don’t confuse g’s with grams! This means it will be 4.22 times heavier than normal. Answer: 1076 g
5) This one is tedious! Use the data for the first part (be sure to subtract weight from the thrust!) to find
that the final speed is 82 m/s and it took 9.4 s to get there. Use the data from the second part to find
that the time in freefall is 21 s. Answer: 30 s
6) The rope can hold all three of them, which is 7g times the mass of Cheetah. To support a stationary
Jane, requires a force of 2gmC, so 5 gmC remains to accelerate Jane. Her maximum acceleration is 5/2
g. Use this and a starting speed of zero and no bothering to slow down at top. Answer: 2.2 s
7) The accelerating force is friction (0.28 mg cos12) which is reduced by the component of gravity that is
parallel to the slope (mg sin12). Answer: 0.65 m/s2
8) This is quite hard. The top block has the greater acceleration because friction and gravity are working
together. For the bottom block: a=8.23, final velocity is –v, and the distance is –A. For the top block
a=3.53, initial velocity is 0, final velocity is v and the distance is 1-A. Set up an equation for the top
block. You can set up another equation if you include t for the bottom block. You can set up a third
one using t for the top block and a=v/t. Solve these three equations. Answer: 4.33/5.33 = 0.81 m.
9) You need to apply F=ma three times. John and Odie before: 93=96a, a = 0.969 m/s2. Odie before: 93F=11a=11(0.969), F =82.3 N. Odie after: a = 82.3/11 Answer: =7.5 m/s2
10) This is like the Slinky – but with 5 discrete masses. The hand is exerting an upward force of 5mg.
When that is removed the top mass has 5 mg down and an acceleration of 5g. Answer: 49 m/s2
Day 7
 Video #1: Friction balances gravity and the normal force makes it go in a circle.
 Video #2: You would go through, continuing at your velocity, tangential to the circle.
 Video #3: At the top gravity provides the net force and so the spring scale reads 0.
 Video #4: The top of the loop is like the previous example and therefore 9.8 = v2/40. At the bottom
you need a lot more force because you will be going faster and because you need to work against
gravity. (More on this in the energy section.)
Draw neat, clear force diagrams that are at least roughly to scale!!
1) The horizontal component of tension provides the net force. The angle can be found using the radius
and length of rope. Answer: 1.3 s
2) The net force is up and = 3(9.8) because one g from the scale is used to cancel gravity. Answer: 85 m
3) The total force is the vector sum of the horizontal frictional force turning the car and the vertical
normal force. Answer: 27 kN.
4) Friction provides the net force. Answer: 24 m/s
5) The net force is the horizontal component of tension. The vertical component of tension plus the
normal force = mg. Answer: 264 N
6) The total force of the road is the vector sum of the horizontal friction causing the speeding up 4 x 1800
and the vertical normal force. The weight minus the normal force is the net force that causes the
circular motion: 1800 (9.8 – 4) Answer: 11 kN.
7) This one is challenging. The net force is limited to the horizontal component of the force of the seat on
the passenger. This can be found because the vertical component cancels gravity. Use this force to find
the radius of the circle (2972 m). This means that when the plane banks 150 m one way and then the
other, the paths are small portions of a circle. Find the angle formed by this arc. (18.3o). This is a
fraction of a circle and this can tell you what fraction of a period it took to turn. (0.102). Find the
period and then the time. Answer: 11 s.
Day 8
1) This question can be done most simply by proportions. You do not need the mass of the Earth – it
stays constant. Use r3/T2 is constant. The new radius will be the old one (the radius of the Earth) times
the cube root of the square of the ratio of the periods. That ratio is 7/3, a simple number because many
things cancel. r = RE (1.759) = RE (1 +0.759). The altitude is the extra part. Answer: 4.8 Mm.
2) All two-body systems orbit about their mutual centre of mass. Usually one of the two objects is much
bigger and that point is almost the same as the centre of the large mass. In this case each star orbits a
point halfway between. This means that the orbital radius (r) is half the separation between the planets
(2r). What if one star had twice the mass of the other? Answer: 2.8 x 107 s for the double star.
3) You need to first find the mass of the planet by d*4/3 r3. (m = 3.67 x 1016 kg.) Answer: 2.9 m/s
4) This question is a great example of why you should simplify your equations before putting values in. If
you solve for T and substitute d4/3 r3 for the mass you will see that the effect of radius cancels out
and the period doesn’t change.
5) At the higher speed the altitude will be smaller. v2 r = GM and will be constant for both situations – so
use proportions. The second radius is 2 r/(2.25) = 0.889 r. This is less than the radius of the planet – so
the orbit is impossible!
6) The acceleration and radius can be used to find the mass of Mars. Answer: 1.71 x 107 m.
7) You need three equations. The net force is spring tension – because gravity will be negligible, so F = k
x. The radius of orbit will be the length of the spring plus the increased length, divided by 2. The force
causes circular motion. Answer: 840 m
8) This question requires a circular motion, then projectile motion and then Pythagorus. The radius and
period give a speed of 10.3 m/s. When the rope breaks it goes into projectile motion, starting with a
horizontal velocity. The vertical information requires a time in the air of 1.2 s. The time and speed
combine to give a horizontal flight of 12.0 m. The swing started 12.8 m from the centre and then went
at right angles. Answer: 17.5 m
9) This question is very similar to the previous one but it also needs a force diagram. The net force is the
horizontal component of tension which results in a speed of 5.42 m/s. The time to fall the height of 22cos(30) is 0.452 s. It travelled 2.45 m during the fall and this must be added to the radius of 1.73
using Pythagorus. Answer: 3.0 m
Day 9
Some of these questions can be done with F=ma. However, it is usually much easier and
sometimes only possible to solve them using conservation of energy.
1) ½ m v2 = Ff d + mgh. Answer: 70 m.
2) The force is always perpendicular to the displacement and so there is no work done.
3) The gravitational energy from dropping 6 m is converted by friction along the slope and then along
the flat. The weight - mg - can be cancelled. 6 = 0.3*cos40*6/sin40 + 0.45 d. Answer: 8.6 m
4) The work done is the increase in gravitational energy for each snow ball. The mass can be found
with density and volume of a sphere and the height is the distance to the centre of the sphere. W =
345*9.8*4/3 * (0.0915 + 0.131 +0.083). Answer: 4.3 kJ
5) ½ m V2 = Ff d + mgh + ½ mv2. However, it runs out of energy before it gets to the top of the slope.
6) The easiest way to do this is to calculate the speed at the bottom, ½ m v2 = mgh. The speed is 24.2
m/s and the average speed on the slopes will be half of this, 12.1 m/s. Then just calculate the three
times using d/v. Answer: 10 s
7) The speed with respect to the ground is the vector addition of the speed of the truck and the
additional speed from throwing. v = 9.18 m/s. Then use conservation of energy, ½ m v2 = Ff d.
Answer: 5.7 m
8) Use conservation of energy to find the speed at the bottom, 18.8 m/s. Then use projectile motion
with the initial velocity being 18.8 m/s 30o above vertical and having to land on a slope 30 o below
horizontal. This is like Jean’s golf game in the projectile motion lesson. The time spent in the air is
3.84 s. Answer: 73 m
9) The car travels 12 m while the driver reacts. The stopping distance with brakes is 45 – 12 = 33 m.
½ m v2 = mgd. Answer: 0.6
Day 10
1) 3.7 m
2) ½ m v2 = Eff (P A d ) Answer: 52 m/s
3) The first situation is static loading and balances forces, so that 3mg = k3. The second situation is
dynamic loading and converts energy, such that mg(h+9) = ½ k x2. Answer: 3.5 m
4) mgh = Eff( V I t). Make sure the hours are changed into seconds. Answer: 84 km
5) The first situation balances forces, (80 +20)g = k x . The second situation converts energy, 20 gh =
½ kx2. Answer: 0.61 m.
6) Use conservation of energy to find that the speed is 14 m/s. Next consider projectile motion. If it
just misses the ceiling then, its vertical velocity is zero at the top and so must have started with a
vertical velocity of 7.67 m/s. Trigonometry and the two velocities means that it was fired 33o
above horizontal. Now you can find the time in air and the horizontal velocity and therefore the
horizontal distance. Answer: 18 m
7) Looking at the geometry shows that when pulled back 1.5 m, the elastic becomes 3.6 m long or 1.6
m longer. It also means that the tension is 1200 N and k = 750 N/m. Use conservation of energy, ½
m v2 = ½ kx2. Answer: 14 m/s
8) For a length of 5.0 m, k = mg/1.5. A shorter length will be stiffer and for a length of 1.0 m, the k is
five times greater or mg/0.3. For any length L, k = mg/0.3 L. Conservation of energy means that
mgh = ½ k x2. When you substitute the known values and the formula for k, you get a quadratic in
L. Answer: 26 m.
9) The velocity at the top will be zero. The velocity at the bottom can be found by conservations of
energy. Answer: 2.8 m/s.
10) Use conservation of energy to find the speed at the bottom, ½ m V2 = mgh + ½ m v2 and V = 24
m/s. Tension balances his weight and also provides the force needed to make him go in a circular.
T = mV2/r + mg = Answer: 6.5 kN
Day 11
The momentum of an isolated system is always conserved.
Kinetic energy is only conserved in an elastic collision.
1) When the man jumps, the cart moves in the opposite direction. The man’s legs determine how fast
he moves relative to the cart – not the ground. If the cart’s velocity relative to the ground is v, then
the man’s velocity must be (15 – v) relative to the ground. Use conservation of P. Answer: 1.0 m/s
2) Use conservation of energy (mgh = ½ mv2) to find the speed at the bottom. Use conservation of
momentum (mV = (m + M) v) to find Cheetah’s speed. Answer: 36 m/s
3) Use conservation of momentum and conservation of kinetic energy to form 2 equations for the first
collision, to find the mother’s speed. Look at the ratio of speeds and use this to find the baby’s
speed without doing the calculations all over again. Answer: 27 m/s
4) Use conservation of momentum to find the speed of Paul’s car. Use conservation of energy (½
mv2= ½ kx2) to find the compression. Answer: 0.84 m
5) Use conservation of momentum to find the speed after the collision. Use conservation of energy (½
mv2 = mgd) to find the distance. Answer: 0.16 m
6) Use 2-D conservation of energy and Pythagoras. Answer: 5/6
7) This can be solved with conservation of P, but is easier if you juts switch into the barge’s frame of
reference. The initial speed is 50 m/s, the final is 0 m/s. You can find the distance if you know the
acceleration. In the Earth frame the plane has an acceleration of F/m = -1/2 g and the barge has an
acceleration of F/m = 1/8 g. What is the plane’s acceleration in he barge frame? Answer: 340 m
8) Use conservation of momentum to find the speeds of Jean and the pepper after throwing. Use
conservation of momentum to find Lucien’s speed after catching. Find their relative speed.
Answer: 12 m/s
9) Find the vertical and horizontal components of the initial velocity. Use the horizontal component
and conservation of momentum to find the horizontal speed of the shot bird. Use the vertical
component to find the time in the air. Answer: 16 m
10) Use conservation of momentum to find the speed after the collision. Use conservation of energy (½
mv2 = mg sin30 d) to find the distance along the slope – not the height. Answer: 23 m
11) At maximum compression, they are not moving closer together not farther apart, so they must be
going at the same speed. Use conservation of momentum to find that speed. Use conservation of
energy (½ mv2= ½ kx2) to find the compression. Answer: 1.1 m
12) Draw the balls just as they collide. The only force is the normal force and this geometry tells you
at what angle the two balls move. Use conservation of momentum in 2-D and conservation of
kinetic energy. Answer: 0.41 m/s back.