PHY131
Ch 2 to 5 Exam
Choose ONLY one problem from Ch 2
An electron moving along the x axis has a position
given by x = 10t e-t m, where t is in seconds. How
far is the electron from the origin when it
momentarily stops?
To find inflection points,
-t
x = 10t e
take derivative and set equal to zero.
-t
d(10t e )/dt;  use product rule
10e-t + 10t(-1)e-t = 0
10e-t = 10te-t
1
= tinflection point
x = 10t e-t
x = 10(1) e-1
x = 3.68 m
16/16
8/8
Name:
An electron has a constant acceleration of
+3.0 m/s2. At a certain instant its velocity
is +9.0 m/s. What is its velocity (a) 2.0 s
earlier and (b) 2.0 s later?
<ONLY 90% max value>
a = vf – vo / t (18/18)
2 sec after to
2 sec before to
3 = vf – 9 / (t+2)
3 = vf – 9 / (t-2)
6 = vf – 9
-6 = vf – 9
15 = vf
3 = vf
(6/6)
(6/6)
9/9
Ch 4 At time t = 0.00, a projectile is launched from ground level. At t = 2.00 s, it is displaced
d = 40.0 m horizontally and h = 60.0 m vertically above the launch point. What is the initial
velocity of the projectile? (b) At the instant it reaches its maximum height above ground level,
what is its horizontal displacement D from the launch point?
vx-ave = Δx / Δt
vx-ave = 40 / 2
vx-ave = 20 m/s
6/6
y = ½ g t 2 + v y t + yo
60 = -5 22 + vy 2 + 0
vy = 40 m/s
18/18
v = 20i + 40j
or
2
2 ½
(20 + 40 ) @ tan-1(40/20)
44.7 m/s
@ 63.4°
As shown force is a Choose ONE problem from Ch 5
function of time
that acts on a 2.00
kg ice block which
only moves along
the x axis. At
t = 0, the block is
moving in the
with a speed of 3.0 m/s. What
positive direction
is its velocity, including sign,
of the axis,
at t = 9 s?
(b)
a = Δv / Δt
10 = 40 / Δt
ttop to bottom = 4 sec
vx-ave = Δx / Δt
20 = Δx / 4
Δx = 80 m
Let the mass of
the block be 4.0 kg
and the angle θ be
30°. Find (a) the
tension in the cord
and (b) the normal
force acting on the
block.
5/5
5/5
(c) If the cord is
cut, find the
magnitude of the
block's acceleration.
F
=m
a
F
= m Δv / Δt
F Δt = m Δv
Area = m Δv
(29-10) = 2 (vf – 3)
vf = 12.5 m/s
6/6
FNet
8/8
= m a
sinθmg – cosθFT = m (0)
sinθmg = cosθFT
FT = sinθmg/cosθ
FT = tanθmg
FT = 23.1 N
6/6
8/8
Count blocks under curve
Positive area (t = 0 to 6 sec)
½ 2(4) + 2(2) + 3(6) + ½ 1(6)
29 Ns
FNormal = cosθmg – sinθFT
FNormal = 46.2 N
Version 1
Negative area (t = 6 to 9)
½(1)(-4) + 2(-4)
-10 Ns
FNet
= m a
sinθmg = m (a)
sinθg = a
5/5
2
a = 5 m/s
Version 2
Negative area (t = 6 to 11)
½(1)(-4) + 2(-4) + ½(2)(-4)
-14 Ns
PHY131
Ch 2 to 5 Exam
Name:
Choose ONLY one from Ch 2
The position function of a particle is x(t) = 8.0 - 3.0 t3 (a) At what
time and (b) where does the particle (momentarily) stop? At what
(c) negative time and (d) positive time does the particle pass
through the origin?
(a)
To find the inflection point
dx
/ dt = 0
3
d(8 – 3t ) / dt = 0
-9t2 = 0
15/15
t = 0 sec
(b)
x(t=0) = 8.0 - 3.0 t3
x = 8 meters
6/6
(c) Not possible
(d)
x(t) = 8.0 - 3.0 t3
0 = 8.0 - 3.0 t3
8.0 = 3.0 t3
t = 1.387
8/8
An electron has a
constant acceleration of
+3.0 m/s2. At a certain
instant its velocity is +9.0
m/s. What is its velocity
(a) 2.0 s earlier and (b)
2.0 s later?
<ONLY 90% max value>
4/4
Ch 4
A ball is shot from the ground into the air. At a height of 10.0 m, its velocity is 8.00 i +
8.00 j m/s. (a) To what maximum height does the ball rise? (b) What is the initial ball's
velocity?
(a)
(b)
a = Δv / Δt
y = ½ a ttop to bottom2
a = vf – vo / t8 to top
13.2 = ½ 10 ttop to bottom2
-10 = 0 – 8 / t8 to top
ttop to bottom = 2.30 sec
t8 to top = 0.8 sec
NOTE:
a = Δv / Δt
Final velocity in the vertical
y = ½ a t 2 + v o t + yo
a = vf – vo / ttop to bottom
direction at the bottom is equal in
2
magnitude and opposite in direction
y = ½(-10)(.8) + 8(.8) + 10
10 = vf – 0 / 2.3
to initial velocity at the start.
y = 13.2 m
vf = 23 m/s
PHY131
Page 2
Choose ONE problem from Ch 5
The figure
gives the force
as a function
of time
component
that acts on a
3.00 kg ice
block that can
move only along
positive direction of the axis, with a
the x axis. At t
speed of 3.0 m/s. What is its
= 0, the block
velocity, including sign, at t = 11 s?
is moving in the
Name:
Let the mass
of the block
be 4.0 kg and
the angle θ be
30°. Find
(a) the tension
in the cord
and
(b) the normal
force acting
on the block.
(c) If the cord is cut,
find the magnitude of
the block's
acceleration.