PHYSICS 231 Review problems for midterm 1 · 2012-10-01 · Review problems for midterm 1 . Some Housekeeping • The 1st exam will be Wednesday October 3. • The exam will take

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.

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PHY 231 4

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 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 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

What is the displacement at t=4 s.

Velo

city

(m

/s)

3 3

1 1

1.5

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

PHY 231 5

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/s 3.30 m/s

6.50 m/s

v?

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)

Cross straight

a)

To cross straight: use picture a)

a) Acceleration in vertical direction is ALWAYS g=–9.81 m/s2; TRUE

b) See a) FALSE c) It is positive going up, negative going down: FALSE d) 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) 9:20 AM B) 9:10 AM C) 9:00 AM D) 8:55 AM E) None, I’m going to BPS 1415 for my exam

PHY 231 8

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 9

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 10

Fg

FgL=Fgcos()

Fg//=Fgsin()

Ffr=kn

n=FgL

1) Draw forces 2) 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=0 18.9-1.12x9.81xsin(36.5)- Ffr=0 Ffr=12.4 N

M T Ffr=n= mg

Fg=mg

n=mg

a) No acceleration: no net force. Ffr=T maximal Ffr=sn so T<= sn : TRUE

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b) k < s and acceleration will start if T> sn so FALSE

c) At rest, so no net force: F=0: FALSE

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.

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14

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

335 km/hr= 93.1 m/s

15

16

L: length units T: time units

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18

19

20

PHY 231 21

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 231 22

PHY 231 23

PHY 231 24

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 25

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 26

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 27

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 28

29

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=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 So a=1.2 m/s2 The acceleration is the same for both masses

1

2

PHY 231 30

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 31

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 32

PHY 231 33

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.

PHY 231 34

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 35

True, see e.g. the drawing

PHY 231 36

30o Tvertical T

200 N

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 37

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=0.5mvi

2=0.5*40*1.52=45 J W=Fx and F=W/x=45/3=15 N Since object was moving to the right and slowing down, force must be pointing to the left.

PHY 231 38

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)

------------------ + Add (1) and (2) -kM1g+M2g=0 So k=M2/M1=2/6=0.33

M1g

n Ffriction

T T

M2g

PHY 231 39

Force in direction of motion: 20cos(60o)=10 N W=Fx=10 x 4 = 40 J

PHY 231 40

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 41

Fg//=mgsin FgL=mgcos

Fg=mg

n=-FgL Ffr

Ffr=sn=smgcos Since the block is not moving: Ffr=Fg// so smgcos=mgsin s= sin/cos= tan=1.15

PHY 231 42

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 43

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 44

False: The component of the velocity that points across the river is largest for boat B, hence he crosses quicker.

PHY 231 45

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 46

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 47

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 48

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 49

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 50

Initial kinetic energy: 0.5mv2=0.5*2000*302=900000 J Work done to stop the car: W=Fx=10000x 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