Department of Physics
PCS 211 – Physics: Mechanics
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Section:
Final Exam – December 10, 2022
Check the name of your professor:
Dr. Da Silva
Dr. Tanguay
Dr. Toronov
Dr. Yuan
Read the following attentively
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This test consists of one full answer problem and 27 multiple-choice questions.
Each question is out of 2 points. In order to score 2 points, you should both select a correct answer
and reasonably justify your selection in the space provided with maximum clarity and detail. If you
only select a correct answer without meaningful justification you can only score 1 point.
For the multiple-choice questions select the option that is closest to the correct answer. Circle it in this
booklet and fill in the appropriate bubble on the bubble sheet. The bubbles must be filled using a
pencil.
This is a closed book test; pens, pencils, erasers, calculators (Sharp EL-546 and the Casio FX-991
only) and your Ryerson University ID are the only allowed items on your desk.
Show all your work in this booklet. We will check your work for all questions, and if your choice is not
justified by the work done on this booklet, credit will not be given.
All electronic devices INCLUDING WATCHES must be turned off and placed out of reach.
Coats, jackets and bags must be placed out of reach (under the seats).
Your TMU photo ID must be on your desk at all times.
Perform ALL your calculations to an accuracy of 5 significant figures and provide your final answer to an
accuracy of 3 significant figures.
Sign here to signify you read the instructions
1
Chapter 5 Newton’s laws, equations of motion, friction; Chapter 6 Particle in uniform
circular motion (Newton's laws)
1. A ball rolls clockwise around the inside of a horizontal pipe. The ball is gradually slowing
down when going from the highest point to the lowest point. At the point shown, with the ball
velocity along the arrow, what is the direction of the net force on the ball?
(A)
(B)
(C)
(D)
(E)
2. Suppose a 1850-kg car is on a roadway that follows the arc of a circle of radius 18.6 m. What
is the maximum speed the car can have without losing contact with the road as it passes this
highest point?
(A) 37.5 km/h
(B) 65.6 km/h
(C) 58.1 km/h
(D) 13.5 km/h
(E) 48.6 km/h
mg = mvmax2/r,
vmax = sqrt(gr)*3.6 = 48.6 km/h
Easy
3. A 5.00- kg box is placed on a scale in an elevator moving downward. What is the scale
reading while the elevator is slowing down to a stop at the acceleration of 3.00 m/s2?
(A) 15.2 N
(B) 49.1 N
(C) 64.0 N
(D) 34.0 N
(E) 77.3 N
N-mg = ma,
N = m(g+a) = 5.00(9.8+3) = 64.0 N
Average
4. A horizontal force of 15.0 N pulls a block of mass 3.90 kg from rest across a level floor. The
coefficient of kinetic friction between the block and the floor is 0.250. If the block is pulled for
12.0 s, determine the final speed of the block.
(A) 25.8 m/s
(B) 37.7 m/s
(C) 46.2 m/s
F-μkmg = ma,
v = at = (F/m- μkg)t = (15/3.9-0.25*9.8)*12= 16.8 m/s
2
(D) 75.6 m/s
(E) 16.8 m/s
Average
5. A 0.200-kg stone is attached to a light spring (with an un-stretched length of 56.0 cm) and
swung in a circle of radius 60.0 cm on a horizontal and frictionless surface. If the stone makes
150 revolutions per minute, find the spring constant.
(A) 49.4 N/m
(B) 740 N/m
(C) 52.9 N/m
(D) 1110 N/m
(E) 880 N/m
ks = mω2r,
k = mω2r/s = 0.200*(150*2π/60)*0.600/(0.600-0.560) = 740 N/m
Average
6. In the Figure below, the mass of the sliding block is 3.00 kg. The mass of the suspended
sphere is 2.00 kg. All surfaces are frictionless. What is the tension in the string connecting the
block and the sphere?
(A) 11.8 N
(B) 19.6 N
(C) 3.92 N
(D) 5.88 N
(E) None of these.
msg - T = msa,
T = mba,
msg - T = msT/mb,
(1 + ms/mb) T = msg,
T = msmbg/(/mb + ms) = 3*2*9.8/(3+2) = 11.76 N
Hard
7. The airplane is moving at 40.0 m/s around a horizontal circle of radius ρ = 270 m. Assume that
the only force other than the gravity applied on the pilot is the normal force from the seat of the plane.
Determine the angle θ the plane is banked at.
(A) 31.2°
(B) 18.6°
(C) 37.2°
mg = N sin θ,
N cos θ = mv2/ ρ,
mg tan θ = mv2/ ρ,
θ = tan-1(v2/ ρg) = tan-1(402/ 270*9.8) = 31.2°
Hard
3
(D) 8.80°
(E) 58.8°
Chapter 7 Work and energy Chapter 8 Conservation of energy and power
8. [conceptual-ish]. A spring which in initially in an un-stretched condition, is first stretched by a length
x and then again by a further length x. How much work is done is done during the second stretch relative
to the first stretch?
a)
b)
c)
d)
e)
W2 = 0.5W1
W2 = 0.33W1
W2 = 2W1
W2 = 3W1
W2 = 4W1
SOLUTION
9. [easy]. A 7500 W engine is propelling a boat at 12 km/h. What force is the water resistance exerting on
the speedboat?
a) 625 N b) 1.6 mN c) 3.3 N d) 2250 N e) 15,000 N
SOLUTION
𝑣 = 12
𝑘𝑚
𝑚
× 1000
× 1ℎ𝑟/3600𝑠 = 3.333 𝑚/𝑠
ℎ
𝑘𝑚
We can then calculate the force the boat would be exerting on the water as:
𝐹=
𝑃
7500 𝑊
=
= 2250 𝑁
𝑣 3.333 𝑚/𝑠
From Newton’s third law, this would have to be the force exerted by the water on the boat, thus is the
water resistance.
4
10. [easy]. A student was cleaning their apartment and began by vacuuming their carpets. The vacuum
cleaner was initially at rest, and they moved it a distance of 45.0 cm by applying a force of 200 N. The
floor was completely flat and horizontal. This resulted in 66 J of work being done on the vacuum.
Determine the angle made between the applied force vector and the floor.
a) 42.8o
b) 37.2o
c) 72.9o
d) -43.0o
e) -90o
SOLUTION
11. [medium]. A 5.00 kg block slides from rest at from a height of h1 = 1.75 m down to a horizontal
surface where it passes over a 2.00 m rough patch. This rough patch has a coefficient of kinetic friction of
0.25 and a coefficient of static friction of 0.44. The block then proceeds to slide up a ramp at an incline of
30.0o what height does the block reach on the inclined ramp (h2) before coming to rest.
a) 1.25 m
b) 0.5 m c) 0.75 m d) 1.75 m e) 3.00 m
SOLUTION
The blocks being and end at rest, thus:
∆𝐸 = 𝑈𝑓 − 𝑈𝑖 = 𝑚𝑔ℎ2 − 𝑚𝑔ℎ1
We have a non-conservative force, friction, which we must determine to establish the work done by
friction. We know that, in the x direction we have −𝑓𝑘 = −𝑚𝑎 and in the y-direction that 𝑁 − 𝑚𝑔 = 0.
We also know that 𝑓𝑘 = 𝜇𝑘 𝑁 = 𝜇𝑘 𝑚𝑔, thus 𝑊𝑓 = −𝜇𝑘 𝑚𝑔∆𝑥. Substituting (where the change in energy
is equal to the work done by friction), we have:
−𝜇𝑘 𝑚𝑔∆𝑥 = 𝑚𝑔ℎ2 − 𝑚𝑔ℎ1
which we can use to solve for h2,
ℎ2 = ℎ1 − 𝜇𝑘 ∆𝑥 = 1.25 𝑚
5
12. [medium]. A 2238-Watt engine pulls a 245 kg block at constant speed up a 12.0 m, 30.0° incline. If
we ignore friction, how long does it take to pull the block up the full length of this incline?
a) 0.74 s
b) 4.22 s
c) 6.44 s
d) 7.01 s
e) 7.85 s
SOLUTION
The work done by the engine in this case would have to equal the work done by the weight. Thus,
𝑊 = 𝑚𝑔ℎ = 𝑚𝑔𝐿 sin 𝜃 = 1.422 × 104 𝐽
From the definition of power,
𝑡=
𝑊 1.442 × 104 𝐽
=
= 6.44 𝑠
𝑃
2238 𝑊
13. [medium]. The driver of a 1,000 kg car travelling at a speed of 16.7 m/s applies the car’s brakes when
he sees a red light. The car’s brakes provide a frictional force of 8,000 N. What is the stopping distance of
the car?
a) 10.4 m b) 17.4 m c) 19.6 m d) 21.0 m e) insufficient information to solve
SOLUTION
1
The car’s kinetic energy is 𝐸𝑘 = 2 𝑚𝑣 2 = 139,445 𝐽. The work done by the case is 𝑊 = (−8000)(∆𝑥𝑜 ).
From the work-energy theorem, ∆𝑥𝑜 = 17.4 𝑚.
6
14. [difficult]. A 2.0-kg block is pushed against a spring of negligible mass and force constant k = 400
N/m, compressing it 0.22 m. The block is then released, and moves along a frictionless, horizontal surface
then up an incline with slope θ = 37.0 (see Figure). How far does the block travel up the inclined plane
before starting to slide back down?
a) 0.544 m
b) 0.821 m c) 0.988 m d) 1.25 m
e) 1.98 m
SOLUTION This solution is too long. This problem is actually simpler.
Let “1” and “2” denote the physical conditions before and after the mass is released, respectively.
Since,
then,
Thus,
And,
This step is not required
If we denote condition “3” to be when the block reaches its highest elevation, then,
Since, v3 = 0, then
where
Thus
And the distance up the ramp would be,
.
7
Chapter 9 Linear momentum and collisions; Centroid, center of mass/gravity;
Chapter 10, 11 Rotational motion, angular speed Torque, angular momentum
15. (conceptual) A student gives a quick push to a ball at the end of a massless, rigid rod, as shown
in the figure, causing the ball to rotate clockwise in a horizontal circle. The rod’s pivot is
frictionless. As the student is pushing, does the torque vector point
16.
a) toward the ball, (b) away from the ball, (c) in the direction of the push, (d) out of the page, or (e)
into the page?
Solution
(e) into the page. The torque direction of the torque can be described using the right-hand rule. If you
orient your right thumb along the radial vector pointing out along the pivot, and your right index finger
along the force, you will find your middle finger points down into the page.
16. (average) An 8.00-cm-diameter hard disk spinning at 7200.0 rpm can stop in 12.0
revolutions. What is the magnitude of the angular acceleration?
Solution
a) 1.65 rad/s2
b) 2.33 rad/s2 c) 3.77 rad/s2 d) 4.92 rad/s2 e) 5.35 rad/s2
f2  i2  2 , and f  i  t . 7200 rpm is
 7200 rot   1 min   2 rad 


  754 rad/s
 1 min   60 s   1 rot 
i  
Rearranging the kinematic equation, and inserting the given numbers, we have
f2  i2  0 rad/s    754 rad/s 

 3.77 103rad/s 2    3.8 103rad/s 2
2
2 12  2 rad 
2

2
8
17. (super easy) An object’s moment of inertia is 2.00 kg m2. Its angular velocity is
increasing at the rate of 4.00 rad/s2. What is the net torque on the object?
a) 1.50 N m
b) 8.00 N m
c) 15.0 N m
d) 24.0 N m
e) 32.0 N m
Solution
  I   (2.0 kg m 2 )(4.0 rad/s 2 )  8.0 kg m 2 /s 2  8.0 N m.
18. (average) A puck attached to a 50.0-cm-long string rotates in a horizontal circle on a
frictionless table at 40.0 rpm. The length of the string is gradually reduced to 25.0 cm as
it is slowly pulled through a hole in the table. What is the puck’s angular velocity, in rpm,
on the shorter string?
a) 10.0 rpm
b) 20.0 rpm
c) 40.0 rpm
d) 80.0 rpm
e)160 rpm
2Solution
Conservation of angular momentum requires that Li  Lf  I ii  I f f . Treating the puck as a particle, we can rewrite
this as
mri 2i  mrf2f
r2
rf
f  i 2 i 
 0.50 m  40 rpm  160 rpm

2 
 0.25 m 
2
19. (average) A 2.00 kg object is moving to the right with a speed of 1.00 m/s when it
experiences an impulse of 4.00 N s, also directed to the right. What are the object’s speed
and direction after the impulse?
a) 1.00 m/s to the right b) 2.00 m/s to the right c) 3.00 m/s to the right d) 4.00 m/s to the right e)
2.00 m/s to the left
Solution
The impulse is equal to the change in momentum, so
9
px  mvfx  mvix  4 Ns
The final velocity is thus
vfx  vix 
4 Ns
4 Ns
 1 m/s 
 3 m/s
m
2 kg
Since the velocity is positive, the object is moving to the right.
20. (average) A 300 g bird flying along at 6.00 m/s sees a 10.0 g insect heading straight
toward it at a speed of 30.0 m/s. The bird opens its beak wide and swallows the bug.
What is the bird’s speed immediately after swallowing?
a) 2.00 m/s
b) 2.50 m/s
c) 3.30 m/s
d) 4.80 m/s
e) 5.20 m/s
Solution
Our system to be bird  bug. This is the case of an inelastic collision because the bird and bug
move together after the collision. Horizontal momentum is conserved because there are no
external forces acting on the system during the collision.
The conservation of momentum equation pfx  pix gives
(m1  m2 )vfx  m1(vix )1  m2 (vix )2
 (300 g  10 g)vfx  (300 g)(60 m/s)  (10 g)( 30 m/s)  vfx  48 m/s
10
21. (average) The three masses shown in FIGURE are connected by massless, rigid rods.
What are the coordinates of the center of mass?
a) x=1.70 cm; y=4.70 cm; b) x=2.50 cm; y=4.20 cm; c) x=3.80 cm; y=1.50 cm;
d) x=0.0 cm; y=10 cm; e) x=5.0 cm; y=5.0 cm;
Solution
xcm 
mA xA  mB xB  mC xC (300 g)(0 cm)  (200 g)(0 cm)  (100 g)(10 cm)

 1.7 cm
mA  mB  mC
(300 g  200 g  100 g)
ycm 
mA yA  mB yB  mC yC (300 g)(0 cm)  (200 g)(10 cm)  (100 g)(8 cm)

 4.7 cm
mA  mB  mC
(300 g  200 g  100 g)
11
Chapter H4 Torque: moment of a force Chapter 12 and H 5 Rigid objects in equilibrium
22. Determine the moment of the force F about the x axis.
a. 95.0 N⋅m
መ ൟN
b. -110 N⋅m
𝐅 = ൛30.0𝒊Ƹ − 20.0𝐣Ƹ + 50.0𝐤
c. -160 N⋅m
d. 210 N⋅m
e. -81.3 N⋅m
2m
𝑧
3m
4m
𝑦
𝑥
23. A uniform ladder 15.0 ft long is leaning against a frictionless wall at an angle of 53.0° above the
horizontal. The weight of the ladder is 30.0 pounds. A 75.0-lb boy climbs 6.00-ft up the ladder.
What is the magnitude of the friction force exerted on the ladder by the floor?
a. 43.0 lb
b. 34.0 lb
c. 38.0 lb
d. 47.0 lb
e. 24.0 lb
12
24. The diagrams below show forces applied to a wheel that weighs 20 N. The symbol W stands for
the weight. In which diagram(s) is(are) the wheel in equilibrium?
a. A
b. B
c. C
d. D
e. A and C
25. Which of the statements below are true about torques and forces acting on a system? Note:
There may be more than one correct answer. Select all that are true.
a) If there is a non-zero net torque about a system’s center of mass, the system’s center of mass
must be accelerating.
13
b) If there is a non-zero net torque about a system’s center of mass, the system must have an
angular acceleration about an axis passing through the center of mass.
c) If the net torque about a system’s center of mass is zero, the net force on the system must also
be zero.
d) Any force acting on a system will create a torque about the system’s center of mass.
26. A uniform beam having a mass of 60 kg and a length of 2.8 m is held in place at its lower end by
a pin. Its upper end leans against a vertical frictionless wall as shown in the figure. What is the
magnitude of the force the pin exerts on the beam?
a. 0.68 kN
b. 0.57 kN
c. 0.74 kN
d. 0.63 kN
e. 0.35 kN
14
̂)N, is acting at 𝐫𝐀 = (10𝐢Ƹ − 4𝐣Ƹ + 7𝐤
መ ) cm. Determine what is the
27. A force 𝐅 = (20𝒊Ƹ − 15𝒋Ƹ + 20𝒌
መ ) cm?
magnitude of the torque that this force produces about the point 𝐫𝐁 = (−4𝐢Ƹ + 5𝐣Ƹ − 8𝐤
a. 93 N⋅cm
b. 58 N⋅cm
c. 120 N⋅cm
d. 77 N⋅cm
15