Transcript SPH4U: Lecture 7 Notes

SPH4U: Practice Problems
Today’s Agenda
Run and Hide
Understanding
An object that’s moving with constant speed travels
once around a circular path. Which of the following
is/are true concerning this motion?
I The displacement is zero
II The average speed is zero
III The acceleration is zero
(a) I only
(b) I and II only
(c) I and III only
(d) III only
(e) II and III only
Understanding
Which of the following is/are true?
If an object’s acceleration is constant, then it
must move in a straight line.
II. If an object’s acceleration is zero, then its speed
must remain constant.
III. If an object’s speed remains constant, then its
acceleration must be zero.
I.
(a) I and II only
(b) I and III only
(c) II only
(d) III only
(e) II and III only
Understanding
A baseball is thrown straight upward. What is the
ball’s acceleration at its highest point?
Gravity is always turned on.
(a) 0
(b) ½ g [down]
(c) g [down]
(d) ½ g [up]
(e) g [up]
Understanding
How long would it take a car, starting from rest and
accelerating uniformly in a straight line at 5 m/s2, to
cover a distance of 200m?
Hint:
 s  v0t 
1
at
2
2
(a) 9.0 s
(b) 10.5 s
(c) 12.0 s
(d) 15.5 s
(e) 20.0 s
1 m  2
 m
200 m   0  t   5 2  t
2 s 
 s 
t 
2
400 m
5
m
s
2
t  8.9 s
Understanding
A rock is dropped off a cliff and strikes the ground
with an impact of 30 m/s. How high was the cliff?
Hint:
We need both the
initial and final
velocity in the
equation
v f  vi  2 a   s 
2
2
v f  vi  2 a   s 
2
2
v f  vi
2
s 
(a) 15.0 m
(b) 20 m
(c) 30 m
(d) 46 m
(e) 60 m
2
2a
2
m

 m
3
0



0 
s 

 s 

m 

2  9 .8 2 
s 

 4 5 .9 m
2
Understanding
A stone is thrown horizontally with an initial speed of 10 m/s from
a bridge. If air resistance could be ignored, how long would it
take the stone to strike the water 80 m below the bridge?
Hint:
We need only think
about the vertical
components.
(a) 1 s
(b) 2 s
(c) 4 s
(d) 6 s
(e) 8 s
 s  v0t 
1
at
2
2
1
m  2
 m
8 0 m   0  t   9 .8 2  t
2
s 
 s 
t 
2
160m
9 .8
m
s
t  4 .0 s
2
Understanding
A soccer ball, at rest on the ground, is kicked with an initial
velocity of 10 m/s at a launch angle of 300. Calculate its total
flight time, assuming air resistance is negligible?
Hint:
We need only think
about the vertical
time to travel up to
its max height, then
double it.
v f  v i  at
t
v f  vi
a
(a) 0.5 s
(b) 1 s
(c) 1.7 s
(d) 2 s
(e) 4 s
m
 m 

0

10
sin
30




 

s
s

 


m 


9.8

2 
s 

 0.51 s
Understanding
A stone is thrown horizontally with an initial speed of 30 m/s from
a bridge. Find the stone’s total speed when it enters the water 4
seconds later?
Hint:
We need only think
about the vertical
time to travel to find
vertical speed, then
combine with
horizontal speed
(a) 30 m/ s
(b) 40 m/s
(c) 50 m/s
(d) 60 m/s
(e) 70 m/s
v f  v i  at
v yf  0
m
 9 .8
s
s
 3 9 .2
v xf  30
m
m
m
s
2
4s
vf 
 v y    vx 
2
2
s

2
m
m


39.2

30




s
s




 49.4
m
s
2
Understanding
Which of the following statements is true
concerning the motion of an ideal projectile
launched at an angle of 450 to the horizontal?
(a) The acceleration vector points opposite to the
velocity vector on the way up and in the same
direction as the velocity vector on the way
down.
(b) The speed at the top of the trajectory is zero.
(c) The object’s total speed remains constant
during the entire flight.
(d) The horizontal speed decreases on the way
down.
(e) The vertical speed decreases on the way up
and increases on the way down.
Understanding
This question concerns the motion of a car on a straight track; the car’s
velocity as a function of time is plotted below
20 velocity (m/s)
10
time (s)
1
2
3
4
5
6
7
−10
−20
(a) Describe what happened to the car at time t=1 s
(b) How does the car’s average velocity between time t=0 sand t=1 s compare to its
average velocity between times t=1 s and t=5 s?
(c) What is the displacement of the car from time t=0 to time t=7s?
(d) Plot the car’s acceleration during this interval as a function of time.
(e) Plot the object’s position during this interval as a function of time. Assuming that
the car begins at s=0.
Understanding
This question concerns the motion of a car on a straight track; the car’s
velocity as a function of time is plotted below
20 velocity (m/s)
10
time (s)
1
2
3
4
5
6
7
−10
−20
(a) Describe what happened to the car at time t=1 s
The car’s velocity remains at 20 m/s, but the acceleration
changes from positive to negative at this time (the foot leaves the
accelerator)
Understanding
This question concerns the motion of a car on a straight track; the car’s velocity
as a function of time is plotted below
20 velocity (m/s)
10
time (s)
1
2
3
4
5
6
7
−10
−20
(b) How does the car’s average velocity between time t=0 sand t=1 s compare to its
average velocity between times t=1 s and t=5 s?
Between t=0 and t=1: the average velocity is ½(0 m/s + 20 m/s)=10 m/s
Between t=1 and t=5: the average velocity is ½(20m/s + 0m/s)=10 m/s
Understanding
This question concerns the motion of a car on a straight track; the car’s
velocity as a function of time is plotted below
20 velocity (m/s)
10
time (s)
1
2
3
4
5
6
7
−10
−20
(c) What is the displacement of the car from time t=0 to time t=7s?
Displacement is the net area between the graph as the time axis
Area=1/2(5*20) – ½(2*10) =40m
Understanding
This question concerns the motion of a car on a straight track; the car’s
velocity as a function of time is plotted below
20 velocity
(m/s) (m/s)
20 velocity
10
10
time (s)
1
1
2
2
3
3
4
5
4
6
5
7 time (s)
6
7
−10
−10
−20
−20
(d) Plot the car’s acceleration during this interval as a function of time.
Understanding
This question concerns the motion of a car on a straight track; the car’s velocity
as a function of time is plotted below
50 velocity (m/s)
20 velocity (m/s)
40
10
time (s)
30
20
10
1
2
3
4
5
6
7
−10
−20
time (s)
1
2
3
4
5
6
(e) Plot the object’s position during this interval as a function of time.
Assuming that the car begins at s=0.
7
Understanding
A cannonball is shot with an initial speed of 50 m/s at a launch angle of
400 toward a castle wall 220m away. The height of the wall is 30 m.
Assume that effects due to the air are negligible.
(a) Show that the cannonball will strike the wall.
(b) How long will it take for the cannonball to strike the wall?
(c) At what height above the base of the wall will the cannonball strike?
Understanding
A cannonball is shot with an initial speed of 50 m/s at a launch angle of
400 toward a castle wall 220m away. The height of the wall is 30 m.
Assume that effects due to the air are negligible.
(a) Show that the cannonball will strike the wall.
R
v sin  2
2
0

g
x  v0 xt
t
x

 251m
We need now only show that the
height of the ball is below 30 m
when the horizontal displacement
is 220m
1
gt
2
2

x
  v 0 sin     
 v cos  
 0
v0 x
2
m

50

 sin  2  40   
s 

R 
m 

9.8

2 
s 

y  v0 yt 
x
v 0 cos  

 x tan    
gx
 1 
x
  g 
  2  v 0 cos  
2
2
2 v 0 cos
  220 m  tan  40   
2
 
m 
2

9.8
220
m



2 
s 

2
m

2
2  50  cos  40  
s 

 23 m
This is less than 30 m, so contact


 
2
Understanding
A cannonball is shot with an initial speed of 50 m/s at a launch angle of
400 toward a castle wall 220m away. The height of the wall is 30 m.
Assume that effects due to the air are negligible.
b) How long will it take for the cannonball to strike the wall?
t
x
v0 x


x
v 0 cos  

220 m
m

50

 cos  40  
s 

 5.7 s
Understanding
A cannonball is shot with an initial speed of 50 m/s at a launch angle of
400 toward a castle wall 220m away. The height of the wall is 30 m.
Assume that effects due to the air are negligible.
c) At what height above the base of the wall will the cannonball strike?
From (a) we have 23 m
Understanding
A physics student is driving home after class. The car is travelling at 14.7 m/s
when it approaches an intersection. The student estimates that he is 20.0 m
from the entrance to the intersection (10.0 m wide) when the traffic light changes
from green to yellow. The light will change from yellow to red in 3.00 seconds.
The maximum safe deceleration of the car is 4.00 m/s2 while the maximum
acceleration of the car is 2.00 m/s2. Should the physics student a) decelerate
and stop or b) accelerate and travel through the intersection? Note: there is a
police car directly behind the student.
Understanding
A physics student is driving home after class. The car is travelling at 14.7 m/s when it
approaches an intersection. The student estimates that he is 20.0 m from the entrance to
the intersection (10.0 m wide) when the traffic light changes from green to yellow. The light
will change from yellow to red in 3.00 seconds. The maximum safe deceleration of the car
is 4.00 m/s2 while the maximum acceleration of the car is 2.00 m/s2. Should the physics
student a) decelerate and stop or b) accelerate and travel through the intersection? Note:
there is a police car directly behind the student.
Let’s try braking first and see how far the car travels in the 3 seconds.
Data Inventory
a   4.0
m
s
2
xi  0 m
v i  14.7
vf  ?
t  3.0 s
xf  ?
We shall use: x f  x i  v i t 
1
at
2
2
m
1
m 
2

x f  0 m   14.7   3.0 s     4 2   3.0 s 
s 
2
s 

m
 0 m  4 4 .1 m  1 8 .0 m
s
 2 6 .1m
Therefore the student will travel 6.1 m into the intersection
(not good with police car behind him)
Understanding
A physics student is driving home after class. The car is travelling at 14.7 m/s when it
approaches an intersection. The student estimates that he is 20.0 m from the entrance to
the intersection (10.0 m wide) when the traffic light changes from green to yellow. The light
will change from yellow to red in 3.00 seconds. The maximum safe deceleration of the car
is 4.00 m/s2 while the maximum acceleration of the car is 2.00 m/s2. Should the physics
student a) decelerate or b) accelerate and travel through the intersection? Note: there is a
police car directly behind the student.
Now let’s try accelerating and see how far the car travels before the light turns red.
Data Inventory
a   2.0
m
s
2
xi  0 m
v i  14.7
vf  ?
t  3.0 s
xf  ?
We shall use: x f  x i  v i t 
1
at
2
2
m
1
m 
2

x f  0 m   14.7   3.0 s     2.0 2   3.0 s 
s 
2
s 

m
 0 m  4 4 .1 m  9 .0 m
s
 5 3 .1m
Therefore the student will travel about 23 m past the
intersection before the light turns red.
Understanding
A car traveling at a constant speed of 30 m/s passes a highway patrol
police car which is at rest. The police officer accelerates at a constant
rate of 3.0 m/s2 and maintains this rate of acceleration until he pulls next
to the speeding car. Assume that the police car starts to move at the
moment the speeder passes the car. Determine:
a) The time required for the police officer to catch the speeder?
b) The distance travelled during the chase?
Understanding
A car traveling at a constant speed of 30 m/s passes a highway patrol police car which is
at rest. The police officer accelerates at a constant rate of 3.0 m/s2 and maintains this rate
of acceleration until he pulls next to the speeding car. Assume that the police car starts to
move at the moment the speeder passes the car. Determine:
a) The time required for the police officer to catch the speeder?
Plan of attack: If we can find a position function for both the motorist
and the police car in terms of time, then we can set both functions
equal to each other (same position) and solve for time
Data Inventory motorist
a0
m
s
x f  x i  vt
2
xi  0 m
vi  3 0
m
v f  30
m
t?
xf  ?
s
s
m

x f  0 m   30  t
s 

m

x f   30  t
s 

Data Inventory Police
a  3.0
m
s
xi  0 m
vi  0
vf  ?
t?
xf  ?
m
s
2
x f  xi  vit 
1
at
2
2
1
m  2
 m
x f  0 m   0  t   3.0 2  t
2
s 
 s 
m  2

x f   1.5 2  t
s 

Understanding
Let’s set then equal and solve
Therefore
m
m  2


30
t

1.5
t



2 
s
s




m
m  2


0   30  t   1.5 2  t
s 
s 



m 
m  
0  t  30   1.5 2  t 
s 
s  

t  0
or
m 

3 0   1 .5 2  t  0
s 
s 
Therefore the police car catches up with
the speeder at 20s (the 0s is when the
car initially passes the police car)
m
m
m 

 1 .5 2  t  3 0
s
s 

30
t
1 .5
m
s
m
s
 20 s
2
Understanding
A car traveling at a constant speed of 30 m/s passes a highway patrol police car which is
at rest. The police officer accelerates at a constant rate of 3.0 m/s2 and maintains this rate
of acceleration until he pulls next to the speeding car. Assume that the police car starts to
move at the moment the speeder passes the car. Determine:
b) The distance travelled during the chase?
Since we know the time (20s) from part a), we need only plug it into the
distance formula from either car. [let’s use the speeder ]
x f  vt
m

  30   20 s 
s 

 600m
Understanding
A stone is thrown vertically upward from the edge of a building 19.6 m
high with an initial velocity of 14.7 m/s. The stone just misses the
building on the way down and strikes the street below. Determine:
a) The time of flight?
b) The velocity of the stone just before it strikes the ground ?
Understanding
A stone is thrown vertically upward from the edge of a building 19.6 m high with
an initial velocity of 14.7 m/s. The stone just misses the building on the way
down and strikes the street below. Determine:
a) The time of flight?
Given:
v i   14.7
a   9.80
m
x f  xi  vit 
Work:
s
0
m
s
1
2
 vi 
t 
t
Relationship: x f  x i  v i t 

1
2
2
2
x f  0m
Required:
at
a t  vit  xi  x f
2
x i  19.6 m
1
v 
Therefore
1 
2
vi  4  a   xi  x f
2 
1 
2 a 
2 
vi  2 a  xi  x f
2

t   1s
t  4s
Flight time is 4 s

a
at
2
2

m

  14.7  
s 

2
m
m 


 14.7   2   9.8 2  19.6 m  0 m 
s 
s 


m
 9.8 2
s
Understanding
A stone is thrown vertically upward from the edge of a building 19.6 m high with
an initial velocity of 14.7 m/s. The stone just misses the building on the way
down and strikes the street below. Determine:
b) The velocity of the stone just before it strikes the ground ?
Given:
v i   14.7
m
a   9.80
m
s
s
2
x i  19.6 m
x f  0m
t  4s
Work:
v f  v i  at
m 
m 

  14.7     9. 8 0 2   4. 0 s 
s  
s 

  2 4 .5
m
s
Required:
Relationship:
vf
v f  v i  at
The negative value in the velocity indicates that
the stone is travelling downward.
Question 1
In a carnival booth, you win a stuffed giraffe if you toss a quarter into a
small dish. The dish is on a shelf above the point where the quarter
leaves your hand and is a horizontal distance 2.1m from your hand. If the
coin is tossed with a velocity of 6.4 m/s at an angle of 600 above the
horizontal, and the coin lands in the dish.
a) What is the height of the shelf above the point where the coin leaves
your hand?
b) What is the vertical component of the velocity of the coin just before it
lands in the dish?
Solution to Question 1
In a carnival booth, you win a stuffed giraffe if you toss a quarter into a small dish. The dish is
on a shelf above the point where the quarter leaves your hand and is a horizontal distance
2.1m from your hand. If the coin is tossed with a velocity of 6.4 m/s at an angle of 600 above
the horizontal, and the coin lands in the dish.
a) What is the height of the shelf above the point where the coin leaves your hand?
b) What is the vertical component of the velocity of the coin just before it lands in the dish?
Since the horizontal speed is
constant, we can determine
how long it took to move
horizontally 2.1m
t
d
v

2.1m
6.4 cos  60  
 0.65625 s
Using this time
and having the
initial point
taken as height
zero, we can
now determine
the height at
t=0.65625s
y  v yt 
 6.4
m
s
 1.5 m
1
gt
2
2
sin  60    0.65625 s  
1
m 
2
9.8
0.65625
s



2 
2
s 
Solution to Question 1
In a carnival booth, you win a stuffed giraffe if you toss a quarter into a small dish. The dish is
on a shelf above the point where the quarter leaves your hand and is a horizontal distance
2.1m from your hand. If the coin is tossed with a velocity of 6.4 m/s at an angle of 600 above
the horizontal, and the coin lands in the dish.
a) What is the height of the shelf above the point where the coin leaves your hand?
b) What is the vertical component of the velocity of the coin just before it lands in the dish?
v yf  v yi  gt
 6.4
m 

sin  60     9.8 2   0.65625 s 
s
s 

m
  0.89
m
s
The negative sign reminds us
that the coin is moving down.
Question 2
Mr. Burns does daredevil stunts in his spare time. His last stunt was to
attempt to jump across a river on a motorcycle. The takeoff ramp was
inclined at 53.00, the river was 40.0 m wide, and the bank was 15.0 m
lower than the top of the ramp. The river was 100m below the ramp.
a) What should his speed have been at the top of the ramp to have
made it to the edge of the far bank?
b) If Mr. Burns’ speed was only half of that in a) where did he land?
Solution to Question 2
Mr. Burns does daredevil stunts in his spare time. His last stunt was to attempt to jump across
a river on a motorcycle. The takeoff ramp was inclined at 53.00, the river was 40.0 m wide,
and the bank was 15.0 m lower than the top of the ramp. The river was 100m below the ramp.
a) What should his speed have been at the top of the ramp to have made it to the edge of
the far bank?
b) If Mr. Burns’ speed was only half of that in a) where did he land?
We first need to determine the
time it takes to be 15m below
starting point and the time to move
horizontally 40m. Then solve these
two simultaneous equations
 1 5  v 0 sin  5 3   t 
1
m  2
9
.8
t

2 
2
s 
4 0 m  v 0 co s  5 3   t
Now
substitute
and solve
for v

 1

40 m
m 
40 m
 15  v 0 sin  53   

9.8
2 
 v cos  53    2 
 v cos  53   
s

0


 0

 15 m  40 m tan  53   
Solving the second
equation for t
t 
40m
v 0 co s  5 3  
v0 
m 

9.8
 800 m 

2 
s 

v
0
cos  53   
2
m 

9.8
 800 m 

2 
s 

15 m  40 m tan  53    cos 2  53  
 17.8
m
s
2
Solution to Question 2
Mr. Burns does daredevil stunts in his spare time. His last stunt was to attempt to jump across
a river on a motorcycle. The takeoff ramp was inclined at 53.00, the river was 40.0 m wide,
and the bank was 15.0 m lower than the top of the ramp. The river was 100m below the ramp.
a) What should his speed have been at the top of the ramp to have made it to the edge of
the far bank?
b) If Mr. Burns’ speed was only half of that in a) where did he land?
Let’s find the time to reach a
vertical position of -100m.
m
1
m  2

 100   8.9  sin  53   t   9.8 2  t
s 
2
s 

t  5.301 s
How about his horizontal
position at the same time
m

x   8.9  cos  53    5.301 s 
s 

 28.4 m