Honors Physics
Fall, 2016
Test Review
Name:
Mr. Leonard
Instructions: The following worksheet is intended to prepare you for the exam on Thursday October 29.
Complete this worksheet and turn it in on the day of the exam.
1. Using graphical methods, find:
~
A
~
B
~+B
~
(a) A
~−B
~
(b) A
~ + 2B
~
(c) A
Honors Physics/Test Review
– Page 2 of 8 –
Name:
2. Use the figure below to answer parts (a)-(d).
~
C
~
A
~
E
~
D
~
B
F~
~
G
~
H
~ + F~ =
(a) E
~ +H
~ =
(b) G
~ −H
~ =
(c) G
~ −G
~ =
(d) H
3. While on a hiking trip, you follow trail A 3.00 miles East. At this point you reach a junction and decide
to take trail B, which goes 5.00 miles 30.0◦ North of West. At this point the trail ends. Not wanting to
retrace your steps, you decide to walk directly back to camp.
(a) Let the x and y directions point East and North respectively. Denote the displacements of trails A
~ and B,
~ and the path you need to take to return to camp as C.
~ Sketch A,
~
and B using the vectors A
~
~
B, and C. Use these sketches as a guide as you complete the rest of the problem.
~ B,
~ and C.
~ Using algebra, solve for C
~ in terms of A
~ and B.
~
(b) Write the vector expression relating A,
~ and B.
~
(c) Determine the x and y-components of A
~ In part (b) we found: C
~ = −A
~ − B.
~
(d) Calculate the x and y-components of C.
~
(e) Calculate the magnitude and direction of C.
m
4. A stone is thrown horizontally from the top of a building. The stone is given an initial velocity of 28.0 .
s
The stone lands 50.0 m from the base of the building. How far above the ground was the stone when it
was thrown?
~v0
∆y = ?
50.0 m
5. A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider
m
is 0.870
at an angle of 35.0◦ above the table, and it lands on the magazine 0.0770 s after leaving the
s
table. Ignore air resistance. How thick is the magazine?
Boris The Spider
∆y = ?
Honors Physics/Test Review
– Page 3 of 8 –
Name:
m
6. A horizontal rifle is fired at a bulls-eye. The muzzle speed of the bullet is 670 . The gun is pointed
s
directly at the center of the bulls-eye, but the bullet strikes the target 0.025 m below the center. What
is the horizontal distance between the end of the rifle and the bulls-eye?
7. In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at
the end. The sharp upward slope launches them into the air, where they perform acrobatic maneuvers.
The end of a launch ramp is directed 63◦ above the horizontal. With this launch angle, a skier attains a
height of 13 m above the end of the ramp. What was the skier’s initial speed?
Honors Physics/Test Review
– Page 4 of 8 –
SOLUTIONS
Solutions to Test Review
1. (a) Remember, we graphically draw the sum of two vectors by placing the vectors head to tail, the draw
an arrow from the tail of the first vector to the head of the last vector.
~
A
~
B
~+B
~
A
~−B
~ is the same as A
~ plus −B,
~ where −B
~ has the same length as B
~ but points in the opposite
1. (b) A
~
~
~ to −B
~ in the usual
direction. Therefore, we flip the direction of B to get −B, then we simply add A
way.
~
−B
~−B
~
A
~
A
~ points in the same direction as B
~ but has twice the length. This means that 2 B
~ points twice as
1. (c) 2 B
far in both the x & y directions.
~
2B
~
A
~ + 2B
~
A
2. (a)
F~
~ + F~ = C
~
E
~
E
2. (b)
~
G
~ +H
~ =D
~
G
~
H
Honors Physics/Test Review
– Page 5 of 8 –
SOLUTIONS
2. (c)
~ −H
~ =A
~
G
~
G
~
−H
2. (d)
~
−G
~
H
~ −G
~ =B
~
H
3. (a)
Bx
30◦
~
B
~
C
By
~
A
3. (b)
~+B
~ +C
~ = ~0
A
~ = −A
~−B
~
C
~ points directly East, so: Ax = 3.00 miles and Ay = 0.00 miles. To find the x and y-components of
3. (c) A
~ we will need to use trigonometry.
B,
Bx = B cos (30◦ )
= −4.33 miles
~ points West.
Note that Bx is negative because B
3. (d)
3. (e)
Cx = −Ax − Bx
= −3.00 miles − (−4.33 miles)
= 1.33 miles
q
~ = Cx2 + Cy2
|C|
q
2
= (1.33 miles) + (−2.5 miles)2
By = B sin (30◦ )
= 2.50 miles
Cy = −Ay − By
= 0.00 miles − (2.50 miles)
= −2.5 miles
Cy
θ = arctan
Cx
= 62.5◦ South of East
= 2.83 miles
4. We start by determining what information was given to us, and what we are asked to solve for.
m
v0,x = 28.0
s
m
∆x = 50.0 m
v0,y = 0.0
s
∆y = ?
m
ay = −9.80 2
s
Essentially what this problem is telling us is that: in the amount of time it takes for the stone to travel
50.0 m horizontally, it falls a vertical distance ∆y. Since there is no acceleration in the x direction, the
Honors Physics/Test Review
– Page 6 of 8 –
SOLUTIONS
m
velocity in the x direction is always 28.0 . Since we know vx and ∆x, we can use this to determine the
s
amount of time the stone was in the air.
∆x
∆t
∆x
∆t =
vx
50.0 m
=
28.0 m
s
= 1.79 s
vx =
Now all we need to do is determine the distance the stone will fall in 1.79 s.
1
2
ay (∆t) + v0,y ∆t
2
1
0
2
*∆t
= ay (∆t) + v0,y
2
1 m
=
−9.80 2 (1.79 s)2
2
s
= −15.6 m
∆y =
The negative sign here indicates that the rock has fallen downwards 15.6 m.
5. As always, we start by determining what information was given to us, and what we are asked to solve
for.
v0 = 0.870
◦
m
s
∆t = 0.0770 s
∆y = ?
θ = 35.0
Since we are given the initial velocity as a magnitude and direction, the first step is to decompose the
velocity into its x and y-components.
~v0
v0,y
◦
35.0
v0,x
v0,x = v0 cos(35◦ )
v0,y = v0 sin(35◦ )
m
m
= 0.713
= 0.499
s
s
We are told that the spider lands on the magazine 0.0700 seconds after jumping. Therefore, the thickness
1
of the magazine is simply ∆y at 0.0700 s. Using ∆y = ay (∆t)2 + v0,y ∆t we can calculate ∆y.
2
1
ay (∆t)2 + v0,y ∆t
2
1
m
m
=
−9.80 2 (0.0770 s)2 + 0.499
(0.0770 s)
2
s
s
= 0.0094 m
∆y =
6. You know the drill...
m
s
m
= 0.0
s
v0,x = 670
v0,y
∆y = 0.025 m
∆x = ?
Honors Physics/Test Review
– Page 7 of 8 –
SOLUTIONS
What this problem tells us is that the time it takes for the bullet to fall 0.025 m is the same as the time
it takes for the bullet to travel some horizontal distance ∆x to the target. Once we know this time, we
1
can use vx = ∆x/∆t to calculate ∆x. We can calculate ∆t using ∆y = ay (∆t)2 + v0,y ∆t.
2
1
0
*∆t
ay (∆t)2 + v0,y
2
2∆y
(∆t)2 =
ay
s
2∆y
∆t =
ay
= 0.071 s
∆y =
Finally, using vx = ∆x/∆t, we can calculate ∆x.
∆x = vx ∆t
m
(0.071 s)
s
= 47.9 m
= 670
7. The known and unknown values are:
θ = 63◦
∆y = 13 m
v0 = ?
The problem asks us to relate the maximum height ∆y = 13 m to the initial velocity. Since the maximum
height is determine by the vertical component of the skier’s motion, first need to relate vy,0 to v0 .
~v0
v0,y
v0,x
Using trigonometry, we can relate v0,y to v0 :
v0,y = v0 sin(63◦ )
m
when a projectile reaches
s
2
2
to top of its trajectory. Plugging this into vf = v0 + 2 a ∆y, we can solve for v0 .
We are told that maximum height is 13 meters. We also know that vy,f = 0
Honors Physics/Test Review
– Page 8 of 8 –
2
2
vf,y
= v0,y
+ 2 ay ∆y
2
*2 0− 2 ay ∆y
v0,y
=
vf,y
2
v0,y
= −2 ay ∆y
2
(v0 sin(63◦ )) = −2 ay ∆y
p
−2 ay ∆y
v0 =
sin(63◦ )
q
−2 9.80 m
s2 (13 m)
=
sin(63◦ )
m
= 17.9
s
SOLUTIONS