Ch4at Time T = 0.00, a Projectile Is Launched from Ground Level. at T = 2.00 S, It Is Displaced

PHY131Ch 2 to 5 ExamName:

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, / 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)
x = 10t e-ttake derivativeand set equal to zero.
d(10t e-t)/dt;  use product rule / 2 sec after to
3 = vf – 9 / (t+2)
6 = vf – 9
15 = vf
(6/6) / 2 sec before to
3 = vf – 9 / (t-2)
-6 = vf – 9
3 = vf
(6/6)
10e-t + 10t(-1)e-t = 0 16/16
10e-t= 10te-t
1 = tinflection point 8/8 / x= 10t e-t
x = 10(1) e-1
x = 3.68 m 9/9

Ch4At 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 t2 + vy t + yo
60 = -5 22 + vy 2 + 0
vy = 40 m/s
18/18 / (b)a = Δv / Δt 5/5
10 = 40 / Δt
ttop to bottom = 4 sec
v = 20i + 40j or
(202 + 402)½@ tan-1(40/20)
44.7 m/s @ 63.4° / vx-ave = Δx / Δt 5/5
20 = Δx / 4
Δx = 80 m
As shownforce is a 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 positive direction of the axis, / Choose ONE problem from Ch 5
with a speed of 3.0 m/s. What is its velocity, including sign, at t = 9 s? / 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.
F= m a
F = m Δv / Δt
F Δt= m Δv
Area = m Δv
(29-10) = 2 (vf – 3)
vf = 12.5 m/s
Count blocks under curve
Positive area (t = 0 to 6 sec)
½ 2(4) + 2(2) + 3(6) + ½ 1(6)
29 Ns
Version 1
Negative area (t = 6 to 9)
½(1)(-4) + 2(-4)
-10 Ns
Version 2
Negative area (t = 6 to 11)
½(1)(-4) + 2(-4) + ½(2)(-4)
-14 Ns / FNet = m a
6/6 8/8
sinθmg – cosθFT = m (0)
sinθmg = cosθFT
FT = sinθmg/cosθ
FT = tanθmg
FT = 23.1 N
6/6 8/8
FNormal = cosθmg – sinθFT
FNormal = 46.2 N
FNet = m a
sinθmg = m (a)
sinθg = a 5/5
a = 5 m/s2 /

PHY131Ch 2 to 5 ExamName:

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) / 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?
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To find the inflection point
d x / dt = 0
d(8 – 3t3) / 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 4/4
(d)
x(t) = 8.0 - 3.0 t3
0 = 8.0 - 3.0 t3
8.0 = 3.0 t3
t = 1.387 8/8

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)
a = Δv / Δt
a = vf – vo / t8 to top
-10 = 0 – 8 / t8 to top
t8 to top = 0.8 sec
y = ½ a t2 + vo t + yo
y = ½(-10)(.8)2 + 8(.8) + 10
y = 13.2 m / (b)
y = ½ a ttop to bottom2
13.2 = ½ 10 ttop to bottom2
ttop to bottom = 2.30 sec
a = Δv / Δt
a = vf – vo / ttop to bottom
10 = vf – 0 / 2.3
vf = 23 m/s / NOTE:
Final velocity in the vertical direction at the bottom is equal in magnitude and opposite in direction to initial velocity at the start.

PHY131Page 2Name:

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 the x axis. At t = 0, the block is moving in the / Choose ONE problem from Ch 5
positive direction of the axis, with a speed of 3.0 m/s. What is its velocity, including sign, at t = 11 s? / 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.