Download Brock physics - Brock University

BROCK UNIVERSITY
Test 2: October 2014
Course: PHYS 1P21/1P91
Examination date: 6 October 2014
Time of Examination: 13:00–13:50
Number of pages: 4 + formula sheet
Number of students: 280
Instructor: S. D’Agostino
No aids are permitted except for a non-programmable, non-graphics calculator. Communications with extraterrestrial and extra-dimensional entities during the test is strictly forbidden.
Solve all problems in the space provided.
Total number of marks: 16
SOLUTIONS
1. [1 point] A soccer ball is kicked along the ground, and it rolls until it stops. The
soccer ball
(a) gradually slows down because the force of the kick gradually decreases as the ball
rolls.
(b) moves at a constant speed until the force of the kick dissipates and the ball
suddenly slows down and stops.
(c) slows down and stops because of Newton’s third law of motion.
(d) gradually slows down because friction opposes its motion.
2. [1 point] A 2-kg ball is thrown from the ground with an initial speed of 10 m/s at an
initial angle of 30◦ . A 1-kg ball is thrown from the ground with the same initial speed
and the same initial angle. If there is no air resistance,
(a) the 2-kg ball goes twice as far, because the gravitational force on it is twice as
large.
(b) the 1-kg ball goes twice as far, because according to Newton’s second law its
acceleration is twice as large.
(c) the two balls have identical paths.
(d) the 2-kg ball goes a bit farther, because it has more momentum.
(e) the 1-kg ball goes a bit farther, because it is a little easier to throw.
3. [1 point] A car is driving at a constant speed on a circular road. The force on a
passenger in the car is
(a) away from the centre of the circle, in the direction of the acceleration.
(b) towards the centre of the circle, in the direction of the acceleration.
(c) zero, by Newton’s first law of motion, because the passenger is not accelerating.
(d) zero, by Newton’s first law of motion, because a body in motion tends to stay in
motion.
For each of the statements below, state clearly whether the statement is true or false.
If the statement is true, briefly explain why, with reference to the relevant laws of
motion. If the statement is false, briefly explain why, with reference to the relevant
laws of motion, and then correct the statement.
4. [2 points] Consider a passenger in a car driving forwards. When the driver slams on
the brakes, the passenger is thrown forward by the force of the car’s motion.
Solution: False. Force is not something that can be carried along with an object, so
there is no such thing as “the force of the car’s motion.”
By Newton’s first law of motion, the passenger tends to continue moving forward at a
constant speed when the car begins to slow down. Relative to the seat, the upper part
of the passenger’s body therefore moves forward a little relative to the lower part of
the passengers’s body. The lower part of the passenger’s body doesn’t move forward
relative to the car because of the strong forces pushing it backwards from the seat belt
and the seat. As the angle of the passenger’s body increases, the lower part of the
passenger’s body exerts increasingly large forces on the upper part of the body (in the
backwards direction) to slow the body down.
As passengers, we feel like we are “thrown forward” because we have a self-centred
perspective; we see the world relative to the reference frame of our bodies. But there is
no force throwing us forward; on the contrary, the forces exerted on us by the seat and
seat belt act in the backward direction, slowing us down. If there was a force acting
on us in the forward direction, we would be catapulted out through the windshield at
a non-healthy speed, because our initial velocity was high, and now there would be
forward acceleration to boot.
All you physics enthusiasts will learn in second year that as the car slows down, the
reference frame of the passenger is non-inertial, and Newton’s laws of motion don’t
apply in the simplest possible way in such reference frames.
5. [2 points] You use your car to push your friend’s car forward, because its engine is
broken. As you slow down together, the force that your car exerts on your friend’s car
is less than the force that your friend’s car exerts on your car.
Solution: False. By Newton’s third law of motion, the force exerted by your car on
your friend’s car has the same magnitude as the force exerted by your friend’s car on
your car.
The state of motion is irrelevant.
6. A baseball is thrown from the edge of a cliff at an angle of 35◦ above the horizontal
with an initial speed of 18 m/s. The edge of the cliff is 12 m above the ground.
(a) [1 point] Determine the time at which the ball reaches maximum altitude.
(b) [1 point] Determine the maximum altitude.
(c) [1 point] Determine the time at which the ball hits the ground.
(d) [1 point] Determine how far from the cliff the ball hits the ground.
(e) [1 point] Determine the speed of the ball as it hits the ground.
Solution: Start by sketching a diagram. On my diagram, I labelled the initial point
of the ball’s path “0,” the point of the path where the ball reaches its peak height “1,”
and the point of the path where the ball hits the ground “2.” I adopted the conventions
that the time at which the ball is thrown is t0 = 0, upwards is the positive direction,
and the height of the ground is y2 = 0.
(a) To determine the time needed to reach the peak altitude, note that the vertical
component of the ball’s velocity at point 1 is zero. Thus,
∆vy
= ay
∆t
∆vy = ay ∆t
∆vy
∆t =
ay
v1y − v0y
t1 − t0 =
−9.8
0 − v0 sin 35◦
t1 − 0 =
−9.8
(18) sin 35◦
t1 =
9.8
t1 = 1.05 s
Thus, it takes a little over one second for the ball to reach its peak altitude.
(b) Use one of the y-displacement equations to determine the maximum altitude. For
example,
2
2
v1y
= v0y
+ 2ay ∆y
2
2
− v0y
v1y
2ay
0 − (v0 sin 35◦ )2
y1 − y0 =
2(−9.8)
(18 sin 35◦ )2
y1 = 12 +
2(9.8)
y1 = 17.44 m
∆y =
The peak altitude above the ground is 17.4 m. If you prefer to calculate the peak
height above the top of the cliff, then the result is
y1 − y0 = 17.4 − 12 = 5.44 m
As a check, use the equation ∆y = v0y t1 + 21 ay t21 to calculate the peak altitude; the
result will be the same as we just calculated.
(c) To calculate the time needed to reach the ground, t2 , make use of the fact that
y2 = 0 when the ball hits the ground. Therefore,
1
∆y = v0y t2 + ay t22
2
1
y2 − y0 = v0 sin 35◦ t2 + (−9.8)t22
2
◦
0 − 12 = 18 sin 35 t2 − 4.9t22
This is a quadratic equation for the time; place it in standard form and then use the
quadratic formula:
4.9t22 − 18 sin 35◦ t2 − 12 = 0
p
(18 sin 35◦ )2 − 4(4.9)(−12)
t2 =
2(4.9)
√
10.3244 + 341.7927
t2 =
9.8
t2 = 2.94 s
18 sin 35◦ ±
The negative solution is rejected because it’s not relevant for this problem. (All you
physics enthusiasts might like to think about its interpretation, though.) Therefore,
the ball hits the ground 2.94 s after it is thrown.
(d) There is no acceleration in the x-direction; thus, the x-component of the ball’s
velocity is constant. Therefore,
∆x = v0x t2
x2 − x0 = v0 cos 35◦ t2
x2 − 0 = (18) cos 35◦ (2.94)
x2 = 43.3 m
Thus, the ball hits the ground 43.3 m from the base of the cliff.
(e) The x-component of the ball’s velocity is constant; therefore,
v2x = v0x = v0 cos 35◦ = 18 cos 35◦ = 14.7447 m/s
The y-component of the ball’s impact velocity can be calculated as follows:
∆vy
∆t
∆vy
v2y − v0y
v2y
v2y
= ay
= ay ∆t
= (−9.8)t2
= (18) sin 35◦ − 9.8(2.94)
= −18.4876
Thus, using Pythagoras’s theorem, the ball’s impact speed is
q
2
2
v2 = v0x
+ v0y
√
v2 = 14.74472 + 18.48762
v2 = 23.6 m/s
Thus, the ball’s impact speed is 23.6 m/s. Notice that the impact speed is greater than
the initial speed. Do you understand why this is so?
7. A ship is 46 km west of a straight shore that runs north-south. The ship moves at a
constant speed of 22 km/h relative to the water at an angle of 21◦ north of east. There
is a steady ocean current straight south with a speed of 11 km/h relative to the shore.
(a) [2 point] Determine the velocity of the ship relative to the shore.
(b) [1 point] Determine how long it takes for the ship to reach the shore.
(c) [1 point] Determine how far south the ship has moved (relative to its starting
position) when it reaches the shore.
Solution: Sketch a diagram!
(a) Let S represent the ship, let G represent the ground (the shore), and let W represent
the water. The basic equation for relative velocity is
~vSG = ~vSW + ~vW G
The given information is
~vSW = (22 cos 21◦ , 22 sin 21◦ )
and
~vW G = (0, −11)
Thus, using the basic equation for relative velocity,
~vSG
~vSG
~vSG
~vSG
= ~vSW + ~vW G
= (22 cos 21◦ , 22 sin 21◦ ) + (0, −11)
= (20.5388, 7.8841) + (0, −11)
= (20.5388, −3.1159)
If you prefer to quote the velocity of the ship in terms of speed and direction:
√
vSG = 20.53882 + 3.11592 = 20.8 km/h
and
3.1159
tan θ =
20.5388
⇒
−1
θ = tan
3.1159
20.5388
= 8.63◦
Thus, the velocity of the ship can also be quoted as 20.8 km/h with direction E8.63◦ S.
(b) The time needed to reach the shore can be calculated by analyzing only the xcomponent of the motion:
∆x = v0x t
∆x
t=
v0x
46
t=
20.5388
t = 2.24 h
The ship takes over 2 hours to reach the shore.
(c) The southward displacement can be determined by analyzing only the y-component
of the motion:
∆y = v0y t
∆y = (−3.1159)(2.24)
∆y = −6.98 km
The ship moves nearly 7 km south relative to its starting position.