CT2-1.
CT2-2.
CT2-3.
CT2-4.
CT2-5.
CT2-6.
CT2-7.
CT2-8.
CT2-9.
CT2-10. A student sees the following question on his CAPA assignment due Friday night
at 10 pm: A ball is thrown vertically upwards with an initial velocity of 21.62 m/s.
Neglecting air resistance, what is the greatest height reached by the ball?
Which of the following formulas should the student use to answer the question?
A)
v = vo + a t
B)
x = x o + v o t + (1/ 2) a t 2
C)
v 2 = v o 2 + 2 a (x - x o )
D)
v =
vo + v
2
Answer: (C) The statement of the problem gives vo, vfinal = v = 0, and a = g. We see the
position. Equation C relates all these quantities. A) and B) are not as useful as C)
because we don't know the time t.
CT2-11. A ball is thrown straight upward. At the top of its trajectory, its acceleration is..
A: zero
B: straight up C: straight down
D: depends on the mass of the ball
Answer: straight down. The acceleration due to gravity is always straight down,
regardless of the velocity of the ball.
CT2-12. A rock is dropped from rest and falls a distance
h (h > 0 ) to the ground. Down is chosen as the positive
direction and the origin is placed at ground level.
What is y0 (the initial position) and what is a (the
acceleration)?
A) y0 = +h , a = – g
B) y0 = –h , a = + g
C) y0 = 0 ,
a=+g
D) y0 = – h , a = – g
E) y0 = 0 ,
a=–g
v0 = 0
h
0
+y
Answer: y0 = –h ,
a=+g
(yfinal = 0)
CT2-13. If you drop an object in the absence of air resistance, it accelerates downward at
9.8m/s2. If instead you throw it downward, its downward acceleration after release is….
A: less than 9.8m/s2.
B: 9.8m/s2
C: more than 9.8m/s2.
Answer: 9.8 m/s2 The acceleration due to gravity is always straight down, always has
magnitude g , regardless of the velocity of the object.
CT2-14. An object's velocity vs. time graph looks like this:
v
t
What situation produces this kind of motion?
A) A rock is thrown straight up.
C) A rock is thrown straight down.
E) None of these.
B) A rock is dropped from rest.
D) A book slides along a table and comes to rest
Answer: A rock is thrown straight up.
CT2-15. On planet X, a cannon ball is fired straight upward. The position and velocity
of the ball at many times are listed below. Note that we have chosen up as the positive
direction.
Time(s)
0
1
2
3
4
5
6
7
8
Height(m)
0
17.5
30
37.5
40
37.5
30
17.5
0
Velocity(m/s)
20
15
10
5
0
-5
-10
-15
-20
y
What is the acceleration due to gravity on Planet X?
A: –5m/s2
D: –20m/s2
Answer: –5m/s2
table.
B: –10m/s2
C: –15m/s2
E: None of these.
Use a = (v2 – v1)/(t2 – t1) and then use any pair of points from the
CT2-15
A ball is fired from a canon straight upward with an initial velocity vo. Assume no air
resistance. If the initial velocity vo is doubled, the time to reach the apex of the
trajectory..
A: doubles.
B: increases by a factor of 4.
C: Neither of these.
D: Impossible to tell from the information given.
Answer: In lecture we derived the formula for the time to reach the apex: t = v0 / g. If v0
is doubled, then t doubles.
If the initial velocity vo is doubled, the maximum height of the ball..
A: doubles.
B: increases by a factor of 4.
C: Neither of these.
D: Impossible to tell from the information given.
Answer: In lecture we derived the formula for the max height of the ball :y = v02 / (2g).
If v0 is doubled, then y increases by a factor of 4.
CT2-17 The distance x traveled by a ball in a time t is given by the formula
x
1
a t 2 where a is a non-zero positive constant.
2
What happens to the distance x
when the time t is tripled?
A) x increases by a factor of 3
C) x increases by a factor of 9
of a
B) x increases by a factor of 9/2
D) x increases by a factor that depends on the value
Answer: x increases by a factor of 9 = 32.