Download PHYSICS 231 Review problems for midterm 1

PHYSICS 231
Review problems for midterm 1
Some Housekeeping
• The 1st exam will be Wednesday October 3.
• The exam will take place right here in BPS 1410.
• We will have assigned seating, so show up early.
• You need to show up 15 min early to guarantee the full 50 min
exam time.
• If you are sick, you MUST have a note from a doctor on the
doctor’s letterhead or prescription pad.
• RCPD forms cannot be turned in for the 1st time on Oct 3.
• You CANNOT use cell phones during the exam.
• You can (should!) bring an equation sheet on a 8.5x11 sheet of
paper.
2
Clicker Quiz!
A) This is an appropriate way to solve physics problems.
B) This will probably not full credit.
3
Velocity (m/s)
v(t)=v(0)+at
x(t)=x(0)+v(0)t+0.5at2
Cut problem up in 1s
pieces
After 1 s:
v(1)=0+0x1=0
x(1)=0+0x1+0.5x0X12=0
1 1
1.5
3
3
After 2 s:
v(2)=v(1)+at=0+3x1=3
x(2)=x(1)+v(1)t+0.5at2
=0+0x1+0.5x3x12=1.5
What is the displacement at t=4 s.
By drawing:
Derive v(t) diagram from a(t) diagram: red line
x(t) is area under v(t) diagram:
PHY 231
After 3 s:
v(3)=v(2)+at=3+2x1=5
x(3)=x(2)+v(2)t+0.5at2
=1.5+3x1+0.5x2x12=5.5
After 4 s:
v(4)=v(3)+at=5-2x1=3
x(4)=x(3)+v(3)t+0.5at2
=5.5+5x1+0.5x(-2)x12=9.5
4
Cross straight
a)
Cross fast
v?
6.50 m/s
3.30 m/s
b)
6.50 m/s
3.30 m/s
3.30 m/s
To cross straight: use picture a)
Velocity of water
Velocity of boat needed to cancel motion of water
Total velocity of the boat (I.e. available in still water)
Velocity ‘left over’ for crossing river
v2+3.302=6.502 so v=(6.502-3.302)=5.6 m/s
Time to cross river: t=x/v=262.0/5.6=46.8 s (use v=x/t)
To cross fast: use picture b)
V=6.5 m/s so t=x/v=40.3 s (but lands downstream)
PHY 231
5
+
a)
b)
c)
d)
Acceleration in vertical direction is ALWAYS g=–9.81 m/s2; TRUE
See a) FALSE
It is positive going up, negative going down: FALSE
It is zero at the start and end and positive everywhere else: TRUE
6
Clicker Quiz!
What time should you arrive at BPS 1410 for your exam
in order to guarantee that you’ll have the full 50 minutes
to take the exam?
A)
B)
C)
D)
E)
9:20 AM
9:10 AM
9:00 AM
8:55 AM
None, I’m going to BPS 1415 for my exam
vx(0)=29.0 m/s
2.19 m
x(t)=x0+v0xt = 0+29t=29t
vx(t)=v0x = 29
y(t)=y0+v0yt-0.5gt2 = 2.19+0t-0.5x9.8t2=2.19-4.9t2
vy(t)=v0y-gt = 0-9.8t=-9.8t
When ball hits ground: y(t)=0 so:
2.19-4.9t2=0 t=0.67s
Use in x(t) equation: x(0.67)=29 x 0.67 = 19.4 m
PHY 231
8
1) Draw forces
2) Since the block is not
moving, no
acceleration, no net
force
mg
T=15N
Note that the tension in the lower block must be trying to pull
The block down-the rope cannot support weight and so not
Produce an upward force by itself…
Tupper-mg-Tlower=0 so Tupper=4.3x9.81+15=42.2+15=57.2 N
PHY 231
9
1) Draw forces
2) The block has
constant velocity,
so no
acceleration, so
no net force!
n=FgL
Fg//=Fgsin()
Ffr=kn
FgL=Fgcos()
Fg
In direction parallel to the slope:
F-Fg//-Ffr=0 so F-Fgsin()-kn=0 and…
F-mgsin()-Ffr=0
18.9-1.12x9.81xsin(36.5)- Ffr=0
Ffr=12.4 N
PHY 231
10
n=mg
Ffr=n= mg
M
T
Fg=mg
a) No acceleration: no net force. Ffr=T
maximal Ffr=sn so T<= sn : TRUE
b) k < s and acceleration will start if T> sn so FALSE
c) At rest, so no net force: F=0: FALSE
11
Clicker Quiz!
When you are passed the stack of exams, what should
you do to identify your exam?
A) Look for the exam with your name and face on it.
B) Make sure the exam doesn’t have someone else’s name
and face on it.
C) Make sure you have only one exam booklet.
D) Make sure you don’t have another exam stuck to the
staple on the back.
E) All of the above.
13
335 km/hr=
93.1 m/s
Work-energy theorem: MEi-MEf=(PE+KE)i-(PE+KE)f=Wnc
14
15
L: length units
T: time units
16
17
18
19
20
225o
1350
Decompose in ‘horizontal’ (west-east direction) and ‘vertical’ (northsouth) components
Horizontal
Vertical
Day 1
15*cos(225)=-10.6
15*sin(225)=-10.6
Day 2
10*cos(135)=-7.07
10*sin(135)=7.07
Sum
-17.7
-3.53
Total displacement:
PHY 231
21
PHY 231
22
PHY 231
23
Initially, the velocity is pointing up, but is decreasing in
magnitude (speed is decreasing) since the gravitational
force is slowing it down. This goes on until it reaches the
highest point, where the velocity/speed equals zero. The
ball than moves down: the velocity becomes negative, but
the speed (not a vector, just a positive number) increases.
So answer c is correct.
PHY 231
24
x(t)=x0+v0t+0.5at2
v(t)=v0+at
X(t)=20-vot-0.5gt2 with g=9.81 m/s2
At t=1.5 s, x=0, so
0=20-1.5vo-0.5*9.81*(1.52)
Solve for vo =-5.98 m/s magnitude: +5.98 m/s answer b)
PHY 231
25
10 km/h

5 km/h
The boat will move under an angle determined by tan=5/10 -> =26.57o
The tan=0.5 also is equal to (distance d)/(width of river=1 km) = d/1
So d=0.5 km.
PHY 231
26
x(t)=x0+v0xt
vx(t)=v0x = v0cos()
y(t)=y0+v0yt-0.5gt2
v(t)=v0y-gt g=9.81 m/s2
v0y=v0sin()
x(t)=x0+v0xt = 40cos(40o)t we don’t now t…
y(t)=y0+v0yt-0.5gt2 = 0.2+40sin(400)t-0.5*9.8*t2 =0 at landing
Solve this for t (quadratic equation): t=5.239 s
Plug this into the equation for x(t):
x(5.239)= 40cos(40o)5.239=161 m answer d)
PHY 231
27
PHY 231
28
1
2
Object 1:
Object 2:
F=m1a, so
F=m2a, so
T-Ffr=m1a (moving to the right)
Fg-T=m2a (note the – sign!!)
m2g-T=m2a
The frictional force Ffr=µkn=µkm1g (magnitude of normal force
equals the gravitational force)
So we have:
Block 1 T-Ffr=m1a -> T-µkm1g=m1a -> T-0.1*40*9.8=40*a
Block 2 m2g-T=m2a ->
-T+10*9.8=10*a
Sum to eliminate T
0-39.2+98=50a
29
So a=1.2 m/s2 The acceleration is the same for both masses
The reading of the scale equals the normal force provided
by the scale.
Write down Newton’s law for the forces acting on you:
F=ma
n-mg=ma (normal force n is pointing up, gravitational force
is pointing down)
The elevator is accelerating upwards, so a>0 and thus:
n>mg which means that the weight you read from the scale
is larger than your nominal weight (i.e. when not
accelerating), answer a)
PHY 231
30
Newton’s law for motion parallel to the slope:
F=ma
-mgsin + Ffan =ma (down the slope is negative)
-0.5*9.81*sin+2.45=0.5*a (cart is at rest, so a=0)
-0.5*9.81*sin+2.45=0, solve for -> =300 answer a)
PHY 231
31
PHY 231
32
Since the skier eventually stops, all the kinetic energy he had at the
bottom of the slope was ‘taken’ (dissipated) by friction. So if we
determine the kinetic energy at the bottom of the slope, we know the
answer to the problem. To find the KE at the bottom of the slope we use
conservation of energy: KEi+PEi=KEf+PEf
KEi : kinetic energy at top = 0 (starts from rest)
PEi : potential energy at top = mgh = 70*9.8*200=137 kJ
PEf : potential energy at bottom = mgh=0 (h=0)
So, KEf = PEi=137 kJ this is also equal to the energy dissipated by
friction after the skier comes to a full stop.
33
PHY 231
The masses are not accelerating since the two masses are
equal. For either mass one can write Newton’s 2nd law:
F=ma=0 T-mg=0 T=mg=1x9.8=9.8 N Answer: False
PHY 231
34
True, see e.g. the drawing
PHY 231
35
T
30o
200 N
Tvertical
The vertical component of the tension in
the left cable must balance the weight of
200 N:
So: Tvertical
Tvertical makes an angle of 90-60=30
degrees with T (total tension in cable):
Cos(30o)=Tvertical/T so
T=Tvertical/cos(30o)=200/cos(300)=231 N
PHY 231
36
Method 1: (positive: motion to the right)
v(t)=v(0)+at at the end v(t)=0 so 0=1.5+at t=-1.5/a
x(t)=x(0)+v(0)t+0.5at2
at the end: x(t)=3 m
So: 3=0+1.5t+0.5at2 use t=-1.5/a
3=-1.52/a+0.5a(-1.5/a)2 3=-2.25/a+1.125/a
And 3=-1.125/a so a=-0.375 m/s2
F=ma = 40x-0.375 = -15 N (-, so the left)
Method 2:
Wnc=Ekin(initial)-Ekin(final) Ekin(final)=0 (at rest) so:
(no change in height, so no change in potential energy)
Wnc=0.5mvi2=0.5*40*1.52=45 J
W=Fx and F=W/x=45/3=15 N
Since object was moving to the right and slowing down, force must be
37
pointing to the left.
PHY 231
n
Ffriction
T
T
M 1g
M2g
Velocity of the blocks is constant, so their acceleration is zero.
For the block on the surface:
F=ma=0
T-Ffriction=0 so T-kn=0 and T-kM1g=0 (1)
For hanging block:
F=ma=0
-T+M2g=0 (2)
------------------kM1g+M2g=0
Add (1) and (2)
So k=M2/M1=2/6=0.33
PHY 231
+
38
Force in direction of motion: 20cos(60o)=10 N
W=Fx=10 x 4 = 40 J
PHY 231
39
1) x(t)=x0+v0xt
68 = 0+35t so t=1.95 s
2) vx(t)=v0x = v0cos()
3) y(t)=y0+v0yt-0.5gt2
4) vy(t)=v0y-gt g=9.81 m/s2 v0y=v0sin()
Use 1) and note that =0o so: 68 = 0+35t so t=1.95 s
Use 3) at t=1.95, where y(t)=0, y(0)=h(eight) solved for:
0=h-0.5 x 9.81 x (1.95)2 so h=18.51
PHY 231
40
Ffr
n=-FgL
Ffr=sn=smgcos
Since the block is not moving:
Ffr=Fg// so
F
=mgsin
g//
FgL=mgcos 
smgcos=mgsin
s= sin/cos= tan=1.15
Fg=mg 
PHY 231
41
Acceleration of the total system: F=ma so T = Ma
where M=1.0+2.0+3.0=6.0 kg and T=35 N, so a=5.8 m/s2
This is also the acceleration of the two blocks pulled by
the string: F=ma Tension = (1.0+2.0) x 5.8=17.5 N
PHY 231
42
1) v(t)=v(0)+at
2) x(t)=x(0)+v(0)t+0.5at2
Use 2) 950 = 0 + 0 +0.5 x 15.5 x t2
To find t=11.07 s for the time it takes to travel 950 m
Use 1) v(t=11.07 s) = 0 + 15.5 x 11.07 = 171.6 m
PHY 231
43
False: The component of the velocity that points across
the river is largest for boat B, hence he crosses quicker.
PHY 231
44
1) v(t)=v(0)+at
2) x(t)=x(0)+v(0)t+0.5at2
Use 2) with x(0)=3 m and v(0)=-6 m/s and a=+5 m/s2
X(t=4) = 3- (6 x 4)+(0.5x5x42) = +19 m
PHY 231
45
1 mile = 5280 ft = 5280 x 12 inch = 5280 x 12 x 2.54 cm=
5280 x 12 x 2.54 x 0.01 m = 1609.3 m
1 m = 1/1609.3 mile = 0.0006213 mile
1 s = 1/3600 hr = 0.000277 hr
331 m/s = 331 x 0.0006213 mile / 0.000277 hr = 742 mile/hr (mph)
PHY 231
46
1) x(t)=x0+v0xt
2) vx(t)=v0x = v0cos()
3) y(t)=y0+v0yt-0.5gt2
4) vy(t)=v0y-gt g=9.81 m/s2 v0y=v0sin()
Use 4) and realize that at the highest point vy=0
0= v0sin()-gt 0=100sin(35)-9.81t t=5.84 s
Use 3) at t=5.84 s
Y(t=5.84) = 0 + 100sin(35)x5.84 -0.5x9.81x5.842=170 m
PHY 231
47
1) v(t)=v(0)+at
2) y(t)=y(0)+v(0)t-0.5gt2
Both rocks undergo 1D motion with a=-g=-9.81 m/s2
Use 2) for rock 1 (upward) and realize that y(t)=0 when the rock lands:
0 = 100 + 15t-0.5x9.81t2 t=6.27 s (quadratic equation, disregard negative
answer)
Use 2 for rock 2 (dropped) to find the time it takes to reach the ground
0=100 +0t -0.5x9.81t2 t=4.515 s.
So if he drops rock 2 6.27-4.52=1.75 s later, they will reach the ground at
the same time
Note: as an alternative, you can calculate the time it takes for rock 1 to
return to the starting height: 100 = 100 + 15t-0.5x9.81t2 you’ll find the
same answer.
PHY 231
48
Mass on surface: F=m1a, so T=15a
(1)
Hanging mass:
F=m2a, so -T+5g=5a (2)
Combine (1) and (2) -T+5g=5T/15
4T/3=5g T=15g/4=15x9.8/4=37N
PHY 231
49
Initial kinetic energy: 0.5mv2=0.5*2000*302=900000 J
Work done to stop the car: W=Fx=10000x
Work done = change in kinetic energy
900000 = 10000 x so x=90 m.
Alternatively, you can use F=ma to calculate the acceleration
And then use the equations
v(t)=v(0)
x(t)=x(0)+v(0)t+0.5at2
To get x(t)-x(0)= x
50
PHY 231