Download Grade 11 Physics – Course Review Part 2

Grade 11 Physics – Course Review Part 2
1. A space traveler whose mass is 75 kg leaves the earth. Calculate his weight:
a. on earth.
b. 400 miles above the earth, where g = 8.1 m/s2.
c. in interplanetary space.
2. Calculate the initial upward acceleration of a 1.3 x 104 kg rocket if the upward
thrust of its engine is 2.6 x 105 N. Do not ignore the weight of the rocket!
3. A hockey puck having a mass of 0.20 kg slides on the ice while slowing down at a
rate of 1.24 m/s2 for 15 m before it stops.
a. If its initial speed was 6.1 m/s, what is the force of friction between the puck
and the ice?
b. What is the coefficient of kinetic friction?
4. A 70 kg crate is dragged across a floor by pulling on a rope.
If the coefficient of static friction is 0.50, what force is required to start the crate in
motion?
a.
b. If the coefficient of kinetic friction is 0.35, what is the acceleration of the
crate caused by the same force?
5. An alpha-particle, the nucleus of a helium atom, has a mass of 6.7 x 10 – 27 kg and a
charge of +2e. What is the magnitude and the direction of the electric field that
will balance its weight?
6. An electric field of 1.92 x 105 N/C is maintained across two plates separated by
1.50 cm. Find the electric force on an oil drop floating between the plates and
carrying a charge of + 50e.
7. If the drop in the above problem has a mass of 1.6 x 10 – 13 kg, what will be its
acceleration as it falls between the two plates? Assume the field points upwards.
8. What sort of field or fields surround a moving electric charge?
9. A wire carrying an electric current experiences no magnetic force as it moves
through a particular region of space. Could there be a magnetic field in this
region? If you think there could be, describe how this could occur.
10.A guitar string is plucked, sending 50 waves along it in 0.200 s. What is
a. the frequency and
b. the period of the waves?
11.Longitudinal pulses of frequency 2.4 Hz are sent along a slinky on the floor at a
speed of 3.6 m/s. What is their wavelength?
12.Three rubber cords, A, B and C, each with a different mass per unit length, are set
up as shown below. The cords are stretched and a pulse is sent along cord A
towards the right.
a. In which of the cords does the pulse travel with the least speed?
b. Is the pulse reflected at the junction of A and B the same way up or
inverted? Sketch the shape of this reflected pulse.
c. When the pulse reflected at the junction of B and C passes back into A, is it
the same way up or inverted?
A
B
C
13.A ray of light in air with an angle of incidence of 35° enters a block of polystyrene.
The angle of refraction is 21°. What is the index of refraction of polystyrene?
14.A train of longitudinal sine waves is sent along a coiled spring having a vibrating
source attached to it. The distance between successive compressions is 25 cm, and
the frequency of the source is 24 Hz. What is the wave speed?
15. The notes most easily produced on a bugle have frequencies of 264 Hz, 352 Hz
and 440 Hz. What is the frequency of the bugle's fundamental note?
Answers:
1. (a) 735N
(b) 607.5N
(c) zero
2. 10.2m/s2
3. (a) 0.248N
(b) 0.13
4. (a) 343N
(b) 1.47 m/s2
5. 2.05 x 10-7 N/C [up]
6. 1.536 x 10-12N
7. 0.20m/s2
8. electric and magnetic fields
9. Yes. The current could be moving parallel to the field lines.
Then FB = IL B sin = 0 because  is zero.
10. (a) 250Hz
(b) 0.0040s
11. 1.5m
12. (a) C
13. 1.60
14. 6.0m/s
15. 88Hz
(b) upright
(c) inverted
Solutions:
1.
Wherever you are, weight is caused by gravity
Fg = m g so:
a) 75 (9.8) = 735 n on Earth b) 75 (8.1) = 607.5 N (a long way from weightless!)
c) zero (since g = 0 far from any planet)
2.
Remember – Newton’s second law tells us acceleration is proportional to the net force on a body.
Fnet = T + Fg = 2.6 x 105 - 1.3 x 10 4 (9.8) = 132 600 N
So a = Fnet ÷ m = 132 600 N ÷ 13 000 kg = 10.2 m/s 2
3.
a) Again find acceleration:
v 2 = vo2 + 2 a d
0 = 6.1 2 + 2a (15)
a = 1.24 m/s 2
Fnet = m a = (0.20)(1.24) = 0.248 N
b) The force in (a) is entirely due to friction, so Fnet = Ff
0.248 =  (0.20)(9.8)
meaning
 = 0.248 / 0.20(9.8) = 0.13
(Note: This coefficient turns out to be equal to a ÷ g. Can you see why?)
Both parts are done by Fnet = Fapplied + Ffriction
a) Here, a = 0 right up till the moment the crate first moves, so 0 = Fapplied + Ffriction
meaning
Fapplied =  s mg = 0.50 (70) (9.8) = 343 N
0.248 =  mg
4.b) This time, the acceleration is not zero, so Fnet = Fapplied + FFriction can be written as
m a = 343 –  k mg
(Recall, we’re using the same applied force)
70 a = 343 – (0.35) (70) (9.8) = 343 – 240.1 = 102.9 N
a = 102.9 ÷ 70 =
1.47 m/s 2
5. Fg = m g = (6.7 x 10 – 27)(9.8) = 6.57 x 10 – 26 N This must be matched by the electric
force, FE = q E
6.57 x 10 – 26 = (3.2 x 10 – 19) E
E = 2.05 x 10 – 7 N/C
Note: This is a very weak field. In the presence of any significant field, it is easy to see
why electric forces completely dominate gravitational forces in the sub-atomic world.
6. FE = q E = (50)(1.6 x 10 – 19)(1.92 x 105) = 1.536 x 10 – 12 N
(The separation of the plates has nothing to do with this force.)
7. The net force on it will be:
Fnet = FE + Fg = 1.536 x 10 – 12 - 1.6 x 10 – 13 (9.8) = - 3.2 x 10 – 14 N (downward)
giving an acceleration of a = Fnet ÷ m = 0.20 m/s2, downward.
8. Both electric fields and magnetics fields surround a moving electric charge.
9. Yes. The current could be moving parallel to the field lines. Then FB = IL B sin = 0
because  is zero.
10. (a) 50 waves in 0.200 s occur at a rate of 250 waves per second. Frequency is 250 Hz
(b) The period is the reciprocal of the frequency, so T = 1/250 s = 0.0040 s
11.λ = v / f = 3.6 / 2.4 = 1.5 m
12.a) C
b) Upright
c) Inverted. The pulse travelled into B as an
upright pulse, then was inverted as it reflected at the boundary between B and C. This
reflected pulse experiences no further inversion as it travels back into A. In fact, the
reflection of a pulse in a low density medium at the boundary of a higher density
medium is the only case in which this inversion occurs.
13. Since n1 sin1 = n2 sin2, we can write 1.00 sin35° = n2 sin21°
n2 = sin35° / sin21° = 1.60 Point-point addition
First, draw the
gives something like
two waves on
this
:  = 0.25 m. So, v = f = (0.25)(24) = 6.0 m/s.
14.
Since
compressions
are
25
cm
apart,
top of each
other.
15.Since all harmonics must be integer multiples of the fundamental, the most reasonable
answer is that these notes are the 3rd , 4th , and 5th harmonics of an instrument that has 88
Hz as its fundamental.
F
I
F
I