PHYSICS 231Review problems for midterm 1

atvv f 0(A) f 0

21 tt( ) 221 attvx o (B)

221 attvx f (C)

tvtvvx f )( 021(D)

f )( 02

)( 221 vvx(E) )( 021 vvx fa (E)

turnonefortimeperiod T

Uniform circular motion with constant speed

turnonefor time period T

radius2 rrv

T

radiansin with arclength rs

tr

tsv

tt

(rad/s)inlocityangular ve2v

l tilt i t)/( 22

(rad/s)in locity angular veTrt

onacceleratilcentripeta)/( 22 rrvac

PHY 2316

m/s

)

3 3

11

Velo

city

(m 3 31.5

v(t) = vo + at

What is the displacement at t=4 s.

x(t) = xo + vot + ½at2

By drawing:By drawing:Derive v(t) diagram from a(t) diagram: red linex(t) is area under v(t) diagram: 9.5 m

7

+

3) Acceleration in vertical direction is ALWAYS g=–9.81 m/s2; TRUE4) See a) FALSE5) It is positive going up, negative going down: FALSE6) It is zero at the start and end and positive everywhere else: TRUE

8

V =8.3

10

1) Draw forces2) The block has

constant velocity

F F i ()

n=Fgy

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

Fgy=Fgcos()

Fgx=Fgsin()

Ffr=kn

Fg

F = 12 4Ffr = 12.4

11

1) Draw forces2) Since the block is not

moving, no acceleration, no net forcemg

T’=?forcemg

T=15N

12

Clicker Quiz!

When you are passed the stack of exams, what should you do to identify your exam?y y y

A) Look for the exam with your name on it.B) Make sure the exam doesn’t have someone else’s name B) Make sure the exam doesn t have someone else s name

on it.C) Make sure you have only one exam booklet.D) M k d ’t h th t k t th D) Make sure you don’t have another exam stuck to the

staple on the back.E) All of the above.)

14

PHY 23115

PHY 23116

Initially, the velocity is pointing up, but is decreasing in magnitude (speed is decreasing) since the gravitational 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 b ll th d th l it b ti b t 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 23117

x(t)=x0+v0t+(1/2)at2

x(t)=20+v t-0 5gt2 with g=9 81 m/s2x(t)=20+vot-0.5gt with g=9.81 m/sat t=1.5 s, x=0, so0=20+1.5vo-0.5*9.81*(1.52)

l f 5 98 / it d 5 98 / b)PHY 231

18

solve for vo =-5.98 m/s magnitude: +5.98 m/s answer b)

x(t)=x0+v0xt v (t)=v = v cos()vx(t)=v0x = v0cos()y(t)=y0+v0yt-0.5gt2

v(t)=v0y-gt g=9.81 m/s2v(t) v0y gt g 9.81 m/sv0y=v0sin()

x(t)=x0+v0 t = 40cos(40o)t we don’t now tx(t)=x0+v0xt = 40cos(40 )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 sSolve th s for t (quadrat c equat on) t 5. 39 sPlug this into the equation for x(t):x(5.239) = 40cos(40o) 5.239 = 161 m (answer d)

PHY 23119

1

Obj t 1 F T F ( i t th i ht)

2

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=m2a2g 2The 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*aBl k 2 T > T 10*9 8 10*

20

Block 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

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 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: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

PHY 23121

accelerating), answer a)

Newton’s law for motion parallel to the slope:F=mamgsin - Ffan =ma (down the slope is positive)0.5*9.81*sin - 2.45 = 0 (cart is at rest, so a=0)solve for -> =300 (answer a)

PHY 23122

solve for > 30 (answer a)

30TverticalT

The vertical component of the tension in the left cable must balance the weight of 200 N:

30oT

Tvertical makes an angle of 90-60=30 degrees with T (total tension in cable):

( 0 ) /

200 N

Cos(30o)=Tvertical/T soT=Tvertical/cos(30o)=200/cos(300)=231 N

PHY 23125

n=-FgLFf

Ffr=sn=smgcos

Fgx=mg sinF =mg cos

Ffr

Since a = 0 (constant speed)Ffr=Fgx so

Fgx mg sinFgy=mg cos

F =mg

gs mg cos = mgs ins= sin/cos= tan=1.15

PHY 23126

Fg=mg

Acceleration of the total system: F=ma so T = Mawhere M = 1.0 + 2.0 + 3.0=6.0 kg and T=35 N, so a=5.8 m/s2where M 1.0 2.0 3.0 6.0 kg and T 35 N, so a 5.8 m/s

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 23127

string F ma Tension (1.0 2.0) x 5.8 17.5 N

atvv f 0(A) f 0

21 tt( ) 221 attvx o (B)

221 attvx f (C)

tvtvvx f )( 021(D)

f )( 02

)( 221 vvx(E) )( 021 vvx fa (E)

1) v(t)=v0+at ) ( ) 02) x(t)=x0 + v0t + (1/2) at2

Use 2) 950 = 0 + 0 +0 5 x 15 5 x t2Use 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

1) ( 11 0 ) 0 15 5 11 0 1 1 6 Use 1) v(t=11.07 s) = 0 + 15.5 x 11.07 = 171.6 m

Or use (E) from the previous slide

PHY 23129

( ) f p

1) x(t)=x0+v0xt 2) vx(t)=v0x = v0cos() 3) (t) t 0 5 t23) 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 sUse 3) at t=5 84 sUse 3) at t=5.84 sY(t=5.84) = 0 + 100sin(35)x5.84 -0.5x9.81x5.842=170 m

O (E) f i lid

PHY 23130

Or use (E) from previous slide

Mass on surface: F=m a so T=15a (1)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/154T/3=5g T = 15g/4 = 15x9.8/4 = 37N

PHY 23131