Download PHYSICS 231 Review problems for midterm 1 1 PHY 231

PHYSICS 231
Review problems for midterm 1
PHY 231
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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
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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)
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+
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
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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
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Acceleration in horizontal direction: use F=ma
You only need the horizontal component, so:
Fx=F x cos()=11.4xcos(21)=10.6 N
a=F/m=10.6/5.22=2.04 m/s2
Now use v=v(0)+at=0+2.04x4.10=8.36 m/s
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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
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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
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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.
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See LON-CAPA
The force pulling M2 to the right is
M2gsin=mgsin
The force pullin M1 down is
M1g=mg
mg>mgsin so M1 wins and M2 goes left,
and M1 goes down (both accelerate).
Also T1=T2
a) Net force on M1: T-mg. M1 goes down, so
gravity wins and M1g>T answer: B
b) Must be equal: connects : answer C
c) T2 wins from M2gsin (M2 goes left) so
answer: B
d) M2g=M1g and M1g>T so M2g>T answer B
e) Equal (else rope breaks)
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Work is equal to the change in potential energy (assume
That kinetic energy at start and stop is 0)
W=mghf-mghi=471x9.81(24.7-0)=114127 J=1.14x105 J
P=W/t=1.14x105/(2.50x60s)=761 J/s = 761 Watt
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335 km/hr=
93.1 m/s
Work-energy theorem: MEi-MEf=(PE+KE)i-(PE+KE)f=Wnc
PEi=mgh = 71.4 x 9.81 x 3490 = 2.44x106 J
KEi=0.5mv2=0.5 x 71.4 x 93.12 = 3.09x105 J
MEi=2.75x106 J
PEf=mgh = 71.4 x 9.81 x 0 = 0 J
KEf=0.5mv2=0.5 x 71.4 x 3.652 = 475 J
MEf=475 J
Wnc= 2.75x106 J – 475 J = 2.75x106 J
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L: length units
T: time units
Write out dimensions:
L = bT3 + cT4
To get length units from the term bT3, b must have units L/T3
To get length units from the term cT3, c must have units L/T4
Answer: C.
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Acceleration is constant, so:
x(t)=x(0)+v(0)t+0.5at2 = 0.5at2
Say the incline is d long.
To cover the full length d:
d=0.5at12 t1=(2d/a)
To cover half the incline (d/2)
d/2=0.5at22 t2=(d/a)
t2/t1 = 1/2 = 0.71 so answer c. It takes more time to cover
the first half (since the average velocity is lower)
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PHY 231
The average speed = distance covered / time that it took
Use:
y(t)=y0+v0yt-0.5gt2 = 19.6t-0.5x9.8t2
vy(t)=v0y-gt = 19.6-9.8t
1) Total distance=twice the distance up
At highest point: vy=0 so 19.6=9.8t t=2
y(2)=19.58 m distance covered: 2x19.58=39.16
2) Total time: when the ball returns: y(t)=0
19.6t-0.5x9.8t2=0 t=0 (start) or t=4
So average speed: 39.2/4=9.8 m/s
Note: you could also have used just the way up….
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Velocity: m/s
a) x2/t units: m2/s not equal to m/s
b) ½ x gt2 units: m/s2 xs2 = m not equal to m/s
c) x/(t-x) units: m/(s-m): this is not possible and
certainly not equal to m/s
d) x/t units: m/s equal to velocity
e) x/t2 units: m/s2 not equal to velocity
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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:
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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.
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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)
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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.
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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)
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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
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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)
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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)
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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.
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PHY 231