BROCK UNIVERSITY
Final Exam: June 2014
Course: PHYS 1P21/1P91
Examination date: 14 June 2014
Time of Examination: 9:00–12:00
Number of
Number of
Number of
Instructor:
pages: 10 (+ formula sheet)
students: 88
hours: 3
S. D’Agostino
A formula sheet is attached at the end of the test paper. No other aids are permitted except
for a non-programmable, non-graphing calculator.
Solve all problems in the space provided.
Total number of marks: 50
SOLUTIONS
1. [5 marks] In a typical greyhound race, a dog accelerates to a speed of 20 m/s over a
distance of 30 m. It then maintains this speed. Determine the greyhound’s time for a
100 m race.
Solution:
v 2 = 2ad
∆t =
=⇒
a=
202
20
v2
=
=
m/s2
2d
2(30)
3
20
70
∆v 70
+
=
+
= 3 + 3.5 = 6.5 s
a
v
20/3 20
2. [5 marks] A driver has a reaction time of 0.50 s, and the maximum deceleration of
her car is 6.0 m/s2 . She is driving at 20 m/s when suddenly she sees an obstacle in the
road 50 m in front of her. Can she stop her car in time to avoid a collision?
Solution: Yes, the driver can avoid a collision:
d = vt +
202
v2
= (20)(0.5) +
= 10 + 33.3 = 43.3 m
2a
2(6)
3. [5 marks] A kangaroo bounds across flat ground, with each jump taking it 10 m from
its takeoff point. The kangaroo leaves the ground at an angle of 20◦ to the horizontal.
Determine (a) the kangaroo’s takeoff speed, (b) the kangaroo’s horizontal speed, and
(c) the time in the air for each jump.
Solution: (a)
1
∆y = 0 = v0 sin 20◦ ∆t − g(∆t)2
2
=⇒
∆t =
2v0 sin 20◦
g
∆x = v0 cos 20◦ ∆t
2v0 sin 20◦
10 = v0 cos 20◦ ·
g
◦
2
v sin 40
10 = 0
g
r
10g
v0 =
sin 40◦
v0 = 12.3 m/s
(b)
v0 cos 20◦ = 11.6 m/s
(c)
∆t =
2v0 sin 20◦
= 0.86 s
g
4. [5 marks] A 23 kg child moves down a straight slide inclined 38◦ above the horizontal.
The child is acted on by his weight, the normal force from the slide, kinetic friction,
and a horizontal rope exerting a force of 30 N as shown in the figure. Determine the
magnitude of the normal force exerted by the slide on the child.
Solution: Draw a free body diagram of the child, and then apply Newton’s second
law to the direction perpendicular to the slide to obtain
N − mg cos 38◦ + 30 sin 38◦
N
N
N
=0
= mg cos 38◦ − 30 sin 38◦
= (23)(9.8) cos 38◦ − 30 sin 38◦
= 159 N
5. [5 marks] The 0.20 kg puck on the frictionless, horizontal table in the figure is connected by a string through a hole in the table to a hanging 1.20 kg block. Determine
the speed with which the puck must move in a circle of radius 0.50 m so that the block
remains hanging at rest.
Solution: The tension in the string required to balance the hanging mass is
T = mg = (1.2)(9.8) = 11.76 N
Therefore, applying Newton’s second law to the puck results in
mv 2
T =
r
and therefore
r
v=
Tr
=
m
r
(11.76)(0.5)
= 5.42 m/s
0.2
6. [5 marks] A 1.5 kg block and a 2.5 kg block are attached to opposite ends of a massless
string. The string hangs over a solid pulley that has diameter 30 cm and mass 0.75
kg. Determine the acceleration of the blocks when they are released. The moment of
inertia of a pulley is 21 mr2 . Assume that the string turns the pulley without slipping,
and that otherwise friction can be ignored.
Solution: Draw free body diagrams for each block and for the pulley. Then apply
Newton’s second law to each block and the pulley, making sure to choose compatible
positive directions for each diagram. You will obtain (after a little algebra in the case
of the pulley) the following three equations:
2.5g − T1 = 2.5a
T1 − T2 = 0.375a
T2 − 1.5g = 1.5a
Solving the first and third equations for the tensions and substituting the resulting
expressions into the second equation results in a single equation for a:
(2.5g − 2.5a) − (1.5g + 1.5a) = 0.375a
g − 4a = 0.375a
4.375a = g
g
a=
4.375
a = 2.24 m/s2
7. [5 marks] A 20 g ball of clay travelling east at 3.0 m/s collides with and sticks to a
30 g ball of clay travelling north at 2.0 m/s. Determine the speed and direction of the
resulting 50 g ball of clay.
Solution: Using the standard coördinate system, applying the principle of conservation of momentum to the collision yields (with all masses in grams)
20(3, 0) + 30(0, 2) = 50~v
Therefore,
20
30
(3, 0) + (0, 2)
50
50
~v = 0.4(3, 0) + (0.6)(0, 2)
~v = (1.2, 0) + (0, 1.2)
~v = (1.2, 1.2)
~v =
Using Pythagoras’s theorem, the magnitude of the final velocity is
√
1.22 + 1.22 = 1.7 m/s
and the direction of the final velocity is 45◦ North of East.
8. [5 marks] A 50 g ice cube can slide without friction up and down a 30◦ slope. The ice
cube is pressed against a spring at the bottom of the slope, compressing the spring by
10 cm. The spring’s stiffness constant is 25 N/m. The ice cube is released. Determine
the distance the ice cube travels up the slope before changing direction.
Solution: The energy initially stored in the compressed spring is
1 2 1
kx = (25)(0.1)2 = 0.125 J
2
2
This energy is first converted into kinetic energy of the ice cube as it moves up the slope,
and then to gravitational potential energy when the ice cube reaches its maximum
height h. Thus,
mgh = 0.125
=⇒
h=
0.125
= 0.255 m
(0.05)(9.8)
The distance travelled by the icecube along the slope is therefore
d=
0.255
= 0.51 m
sin 30◦
9. [10 marks] Circle the best response in each case.
(a) You stand on a bathroom scale in a moving elevator, when the cable suddenly
breaks. The scale reading
i.
ii.
iii.
iv.
v.
increases somewhat.
decreases somewhat.
does not change.
decreases suddenly to zero.
[It depends on the initial speed and direction of the elevator.]
(b) You simultaneously release two balls from a height; one you drop straight down,
and the other you throw horizontally. Which ball reaches the ground first?
i. The ball dropped straight down reaches the ground first, because it travels a
shorter distance.
ii. The ball thrown horizontally reaches the ground first, because its speed is
greater.
iii. The two balls reach the ground at the same time, because their vertical motion
is the same.
iv. [It depends on the masses of the balls; the heavier ball falls faster.]
v. [It depends on the masses of the balls; the heavier ball falls more slowly.]
(c) Why is it difficult for a car to negotiate a curve at high speed?
i.
ii.
iii.
iv.
A huge force is pushing the car outward.
The sliding friction force is too large.
The faster the car moves, the harder it is for the driver to steer.
The magnitude of the friction force may not be large enough to provide the
necessary acceleration.
v. Because of Newton’s third law, the car’s reaction force may be too great.
(d) How would the gravitational force exerted by the Sun on the Earth change if the
Sun suddenly shrank to half of its size, while its mass remained the same. The
force would
i.
ii.
iii.
iv.
v.
increase by a factor of 2.
increase by a factor of 4.
decrease by a factor of 2.
decrease by a factor of 4.
remain unchanged.
(e) The centre of mass of a washer that has constant density
i.
ii.
iii.
iv.
v.
is at its geometrical centre.
is spread throughout the mid-line of the washer.
does not exist, because the centre of the washer is not on the washer.
does not exist, because the washer is not solid.
[There is not enough information given.]
(f) You are at the back of a sailboat and have an electric fan and a source of electric
power. To get the sailboat moving, you should aim the fan
i.
ii.
iii.
iv.
at the sail, so that the wind from the fan will push the sail.
out the back of the boat, so that the boat will move forward.
[There is no way to get the boat moving with a fan.]
[There is not enough information given.]
(g) An object slides in a straight line on a frictionless surface. In order to continue
its motion at a constant speed,
i.
ii.
iii.
iv.
v.
a constant pushing force is needed.
a steadily increasing pushing force is needed.
a steadily decreasing pushing force is needed.
no external force is needed.
no external force is needed provided that the object contains its own force.
(h) Two cylinders have the same dimensions and the same mass, but Cylinder A
is solid and Cylinder B is hollow. (They are made of different materials.) The
moment of inertia of Cylinder A
i.
ii.
iii.
iv.
v.
is greater than the moment of inertia of Cylinder B.
is less than the moment of inertia of Cylinder B.
is the same as the moment of inertia of Cylinder B.
[It depends how large the hollow is.]
[It depends on the specific materials that the cylinders are made of.]
(i) You pick up a book that is lying on the floor and place it on a desk. The total
mechanical energy of the book
i.
ii.
iii.
iv.
v.
has increased, because its potential energy has increased.
has decreased, because it is farther from the centre of the Earth.
remains the same, because it is still at rest.
remains the same, because no work is done on the book.
remains the same, because the kinetic energy of the book does not change.
(j) Ball A is dropped, and a short time later Ball B is dropped. When Ball B is
dropped, the distance between the balls is 5 cm. As the balls fall for a few more
seconds, and before they reach the ground, the distance between the balls
i.
ii.
iii.
iv.
v.
increases.
decreases.
remains the same.
[It depends on the initial height.]
[It depends on how long the balls fall.]
Answers:
(a) iv
(b) iii
(c) iv
(d) v
(e) i
(f) ii
(g) iv
(h) ii
(i) i
(j) i