Download AP Physics Chapter 2 Review

AP Physics
Chapter 2
Review
1. The position of a particle moving along the x axis is given by
x = (21 + 22t - 6.0 t2) m, where t is in seconds. What is the
average velocity during the time interval t = 1.0 sec to t = 3.0
sec?
Δx [(21+ 66 − 54 ) − (21+ 22 − 6)]m
m
v= =
= −2 sec
Δt
2sec
€
dv
v=
=
dt
d (21+ 22t − 6t 2 )
dt
= 22 −12t
m
v1sec + v 3 sec (10 −14 ) sec
m
v=
=
= −2 sec
2
2
2
2. A bullet is fired through a board, 14.0 cm thick, with its line of
motion perpendicular to the face of the board. If it enters with
m
m
a speed of 450 sec
and emerges with a speed of 220 sec
, what is
the bullet's acceleration as it passes through the board?
€
2
f
2
i
2
f
2
i
v = v + 2aΔx
€
v −v
=a
2Δx
m 2
sec
m 2
sec
(220 ) − (450 )
2(0.14m )
=a
−550357 secm 2 = a
3
3. The position of a particle moving along the x axis is given by
x = 6.0t2 - 1.0t3, where x is in meters and t in seconds. What
is the position of the particle when it achieves its maximum
speed in the positive x direction?
dv
2
v=
=12t − 3t
dt
dv
a=
=12 − 6t
dt
At max v, set a = 0.
€
2
x = 6(2) − 2
3
x = 16m
t = 2sec
4
€
4.
A particle moving along the x axis has a position given by
x = (24t - 2.0t3) m, where t is measured in seconds. How far
is the particle from the origin (x = 0) when the particle is
not moving?
dx
2
v=
= 24 − 6t
dt
2
0 = 24 − 6t
t = 2sec
x = (24t − 2t 3 ) m
x = 48m −16m
x = 32m
€
€
5
5. Vx is the velocity of a particle moving along the x axis as
shown below. If x = 2.0 m at t = 1.0 sec, what is the position
of the particle at t = 6.0 sec?
v( )
m
sec
€
t(sec)
At t = 1 sec, x = 2m.
€
Find the area under the line from t = 1 sec to t = 6 sec
and add the results to x = 2m
6
v( )
m
sec
€
t(sec)
€
The two red areas cancel out
and the green area = -3m.
2m − 3m = −1m
7
6. At t = 0, a particle is
m
a
(
located at x = 25m and has sec
m
a velocity of 15 sec
in the
positive x direction. The
€
acceleration of the particle
varies with time as shown
in the€diagram above.
What is the velocity of the
particle at t = 5.0 sec?
2
)
Area is equal to the ∆v.
( )
Δv = 12 6 secm 2 (5sec) = 15 secm
m
m
v 5sec =15 sec
+15 sec
= 30 secm
8
€
7. A particle confined to motion along the x axis moves with
constant acceleration from x = 2.0 m to x = 8.0 meters during
a 2.5 second time interval. The velocity of the particle at x =
m
8.0 m is 2.8 sec
. What is the acceleration during this time
interval?
Δd
6m
m
v= € =
= 2.4 sec
Δt 2.5sec
vi + v f
m
m
v=
v f − v i 2.8 sec
− 2 sec
2
a=
=
t
2.5sec
2v − v f = v i
m
m
2(2.4 sec
−
2.8
)
sec = v i
m
v i = 2 sec
a = 0.32 secm 2
9
€
8. An automobile moving along a straight track changes its
m
m
velocity from 40 sec
to 80 sec
in a distance of 200 m. What is
the (constant) acceleration of the vehicle during this time?
€
2
€f
2
i
v = v + 2aΔx
v 2f − v i2
a=
2Δx
m 2
sec
m 2
sec
80 ) − (40 )
(
a=
2(200m)
a = 12 secm 2
10
€
9. In 2.0 seconds, a particle moving with constant acceleration
along the x axis goes from x = 10 m to x = 50 meters. The
m
velocity at the end of this time interval is 10 sec
. What is the
acceleration of the particle?
1
2
Δx = v f t − at
2(Δx − v f t )
2
2
€
=a
t
m
2(40m − (10 sec
)(2sec))
(2sec)
2
=a
a = −10 secm 2
11
€
10. An electron, starting from rest and moving with a constant
acceleration, travels 2.0 cm in 5.0 msec. What is the
magnitude of this acceleration?
1
2
Δx = at
2
2Δx
=a
2
t
2(0.02m )
(0.005sec)
2
=a
a = 1600 secm 2
12
11. A rocket, initially at rest, is fired vertically with an upward
acceleration of 10 secm . At an altitude of 0.50 km, the engine
of the rocket cuts off. What is the maximum altitude it
achieves?
2
€
vo = 0
€
v Highest = 0
We’re going to find the velocity of the rocket
when the engine runs out of fuel (quits), first.
Then we will find the height the rocket achieves
when moving under only the influence of gravity.
2
f
€
2
i
v = v + 2aΔy
2
f
v = 2aΔy
v f = 2aΔy
vo = 0
€
(
)
v f = 2 10 secm 2 (500m)
m
v f = 100 sec
14
Now we’ll find distance B and then add
distance B to distance A.
v Highest = 0
B
€
€
v 2f = v i2 + 2gΔy
0 = v i2 + 2gΔy
v =100 secm
€
A
−(100
(
m 2
sec
2 −9.8
vo = 0
€
H = Δy A + Δy B
H =1010m
−v i2
= Δy B
2g
)
m
sec 2
)
= Δy B
Δy B = 510m
€
€
15
€
12. A ball is thrown vertically upward with an initial speed of
m
20 sec
. Two seconds later, a stone is thrown vertically (from
the same initial height as the ball) with an initial speed of
m
24 sec
. At what height above the release point will the ball
and stone pass each other?
€
€
The ball, after two seconds:
(
)
v f = v i + gt = 20 secm + −10 secm 2 (2sec) = 0
Now, the ball is starting down with an initial
velocity of zero while the stone is starting up.
€
H ball = 20m
16
1
2
20 − H s = gt
20 − 12 gt 2 = 24t − 12 gt 2
2
20 = 24t
×
20m
€
1
2
H s = 24t − gt
H s = 24t − 12 gt 2
€
€
t = 0.83sec
2
€
€
( ) (
1
2
H s = 24 0.83 − 10
m
sec2
)(0.83)
2
H s = 16.53m
€
17
13. An object is thrown vertically and has an
m
upward velocity of 18 sec
when it reaches
one fourth of its maximum height above its
launch point. What is the initial (launch)
speed of the object?
€
vTop = 0
3H
4
€
€
v 1 = 18 secm
4
vo
€
v 2f = v 21 + 2g( 43 H )
4
0 = v 21 + 2g( 43 H )
4
0 = v 21 + g( 32 H )
4
( ) = H = 22m
2
2 v1
4
3g
€
18
13.2
v 2f = v o2 + 2g(H )
0 = v o2 + 2g( H )
v o = 2gH
(
)
v o = 2 9.8 secm 2 (22m)
m
v o = 20.7 sec
€
19
14. A stone is thrown from the top of a building with an initial
m
velocity of 20 sec
downward. The top of the building is 60 m
above the ground. How much time elapses between the
instant of release and the instant of impact with the ground?
€
4.9t 2 − v o t ± Δy = 0
4.9t 2 + 20t − 60 = 0
2
t=
€
−20 + 20 − 4 (4.9)(−60)
9.8
t = 2.01sec
14
20
€
15. An object is thrown downward
with an initial (t = 0) speed of
2
−(60 − H s ) = −10t − 4.9t 10 from a height of 60 m
above the ground. At the same
×
€ instant (t = 0), a second object
is propelled vertically upward
H s = 40t − 4.9t 2
from ground level with a speed
of 40 . At what height above
the ground will the two objects
pass
each other?
€
m
sec
60m
€
€
m
sec
€
2
60 −10t − 4.9t = 40t − 4.9t
60 = 50t
t = 1.2sec
€
2
Hs = 40(1.2) − 4.9(1.2)
2
H = 40.944m
21
16. A rock is thrown downward from an unknown height
m
above the ground with an initial speed of 10 sec
. It strikes the
ground 3.0 seconds later. Determine the initial height of the
rock above the ground.
€2
−Δy = −10t − 4.9t
Δy =10(3sec) + 4.9( 3sec)
2
Δy = 74.1m
16 €
22
17. A ball thrown vertically from ground level is caught 3.0
seconds later by a person on a balcony which is 14 meters
above the ground. Determine the initial speed of the ball.
Δy = v o t − 4.9t 2
Δy + 4.9t 2
= vo
t
14m + 4.9
m
sec 2
3sec
m
v o = 19.36 sec
€
( 3sec)
2
= vo
23
vTop = 0
1
3
€
H
18. An object is thrown vertically upward such
m
that it has a speed of 25 sec
when it reaches
two thirds of its maximum height above the
launch point. Determine this maximum
height.
€
v 2 = 25 secm
€
3
v 2f = v 22 + 2g( 13 H )
3
2
3
€
€
vo
H
2
0 = v 2 + 2g( 13 H )
3
( ) = H = 95.7m
3 v 22
3
2g
€
24