PHY 2311

PHYSICS 231Review problems for midterm 1

PHY 23120

v(t)=v(0)+atx(t)=x(0)+v(0)t+0.5at2

Cut problem up in 1s piecesAfter 1 s:v(1)=0+0x1=0x(1)=0+0x1+0.5x0X12=0

After 2 s:v(2)=v(1)+at=0+3x1=3x(2)=x(1)+v(1)t+0.5at2

=0+0x1+0.5x3x12=1.5

After 3 s:v(3)=v(2)+at=3+2x1=5x(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=3x(4)=x(3)+v(3)t+0.5at2

=5.5+5x1+0.5x(-2)x12=9.5

What is the displacement at t=4 s.

Velo

city

(m

/s)

3 3

11

1.5

By drawing:Derive v(t) diagram from a(t) diagram: red linex(t) is area under v(t) diagram:

PHY 23121

Cross fast

3.30 m/s

6.50 m/s

To cross fast: use picture b)V=6.5 m/s so t=x/v=40.3 s (but lands downstream)

b)

3.30 m/s3.30 m/s

6.50 m/s

v?

Velocity of waterVelocity of boat needed to cancel motion of waterTotal 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/sTime to cross river: t=x/v=262.0/5.6=46.8 s (use v=x/t)

Cross straight

a)

To cross straight: use picture a)

PHY 23122

a) Acceleration in vertical direction is ALWAYS g=–9.81 m/s2; TRUEb) See a) FALSEc) It is positive going up, negative going down: FALSEd) It is zero at the start and end and positive everywhere else: TRUE

+

PHY 23123

vx(0)=29.0 m/s

2.19 m

x(t)=x0+v0xt = 0+29t=29tvx(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=-9.8tWhen ball hits ground: y(t)=0 so:2.19-4.9t2=0 t=0.67sUse in x(t) equation: x(0.67)=29 x 0.67 = 19.4 m

PHY 23124

Acceleration in horizontal direction: use F=maYou only need the horizontal component, so:Fx=F x cos()=11.4xcos(21)=10.6 Na=F/m=10.6/5.22=2.04 m/s2

Now use v=v(0)+at=0+2.04x4.10=8.36 m/s

PHY 23125

Fg

FgL=Fgcos()

Fg//=Fgsin()

Ffr=kn

n=FgL

1) Draw forces2) The block has

constant velocity, so no acceleration, so no net force!

In direction parallel to the slope:F-Fg//-Ffr=0 so F-Fgsin()-kn=0 and…F-mgsin()-Ffr=018.9-1.12x9.81xsin(36.5)- Ffr=0Ffr=12.4 N

PHY 23126

1) Draw forces2) Since the block is not

moving, no acceleration, no net forcemg

T=15N

Note that the tension in the lower block must be trying to pullThe block down-the rope cannot support weight and so notProduce an upward force by itself…Tupper-mg-Tlower=0 so Tupper=4.3x9.81+15=42.2+15=57.2 N

PHY 23127

MTFfr=n= mg

Fg=mg

n=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 FALSEc) At rest, so no net force: F=0: false.

PHY 23128

See LON-CAPA

The force pulling M2 to the right isM2gsin=mgsinThe force pullin M1 down isM1g=mgmg>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, sogravity wins and M1g>T answer: B

b) Must be equal: connects : answer Cc) T2 wins from M2gsin (M2 goes left) so

answer: Bd) M2g=M1g and M1g>T so M2g>T answer Be) Equal (else rope breaks)

PHY 23129

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

PHY 23130

Work-energy theorem: MEi-MEf=(PE+KE)i-(PE+KE)f=Wnc

PEi=mgh = 71.4 x 9.81 x 3490 = 2.44x106 JKEi=0.5mv2=0.5 x 71.4 x 93.12 = 3.09x105 JMEi=2.75x106 J

PEf=mgh = 71.4 x 9.81 x 0 = 0 JKEf=0.5mv2=0.5 x 71.4 x 3.652 = 475 JMEf=475 J

Wnc= 2.75x106 J – 475 J = 2.75x106 J

335 km/hr=93.1 m/s

PHY 23131

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.

L: length unitsT: time units

PHY 23132

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=0.5at1

2 t1=(2d/a)

To cover half the incline (d/2)d/2=0.5at2

2 t2=(d/a)

t2/t1 = 1/2 = 0.71 so answer c. It takes more time to coverthe first half (since the average velocity is lower)

PHY 23133

The average speed = distance covered / time that it tookUse:

y(t)=y0+v0yt-0.5gt2 = 19.6t-0.5x9.8t2

vy(t)=v0y-gt = 19.6-9.8t1) Total distance=twice the distance up

At highest point: vy=0 so 19.6=9.8t t=2y(2)=19.58 m distance covered: 2x19.58=39.16

2) Total time: when the ball returns: y(t)=019.6t-0.5x9.8t2=0 t=0 (start) or t=4So average speed: 39.2/4=9.8 m/s

Note: you could also have used just the way up….

PHY 23134

Velocity: m/sa) x2/t units: m2/s not equal to m/sb) ½ x gt2 units: m/s2 xs2 = m not equal to m/sc) x/(t-x) units: m/(s-m): this is not possible and

certainly not equal to m/sd) x/t units: m/s equal to velocitye) x/t2 units: m/s2 not equal to velocity

PHY 23135

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

Decompose in ‘horizontal’ (west-east direction) and ‘vertical’ (north-south) components

225o

1350

Total displacement:

PHY 23136

PHY 23137

PHY 23138

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 23139

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, so0=20-1.5vo-0.5*9.81*(1.52)Solve for vo =-5.98 m/s magnitude: +5.98 m/s answer b)

PHY 23140

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/1So d=0.5 km.

PHY 23141

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 landingSolve this for t (quadratic equation): t=5.239 sPlug this into the equation for x(t):x(5.239)= 40cos(40o)5.239=161 m answer d)

PHY 23142

43

Object 1: F=m1a, so T-Ffr=m1a (moving to the right)Object 2: F=m2a, so Fg-T=m2a (note the – sign!!)

m2g-T=m2aThe 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*aBlock 2 m2g-T=m2a -> -T+10*9.8=10*aSum to eliminate T 0-39.2+98=50aSo a=1.2 m/s2 The acceleration is the same for both masses

1

2

PHY 23144

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 23145

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 23146

PHY 23147

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.