Name _______________________

Physics 201 practice exam 3 Page 1

Physics 201 Third Exam—Practice w/answers

Equations:x = x0 + v0·t + ½a·t2v = v0 + a·tv2 = v0

2 + 2a(x-x0)v = (v – v0)/2

v = ∆x/∆ta = ∆v/∆tar = v2/r

vx = v·cos(θ)vy = v·sin(θ)v2 = vx

2 + vy2

tan(θ) = vy/vx

ΣF = maF12 = - F21F(G) = mgF(f)k = µkFNF(f)s ≤ µsFN

p = mvF = ∆p/∆tImpulse = F∆t = ∆pmv1+mv2=mv1’+mv2’

W = Fd·cos(θ)KEt = ½mv2

GPE = mgyKE0 + PE0 + W = KEf + PEfSPE = ½kx2

1. Define the following quanties:a) Work

b) Impulse

c) Potential Energy

d) Momentum

2. In which of the following is work done (by the physics definition) on the object?Explain why or why not.

a) You push hard on a cement wall

No work since the wall does not move: no distance.

b) You carry your lunch tray across the cafeteria at the same height.

No work since the force is perpendicular to the direction of motion.

c) You draw a circle on a sheet of paper with a pencil

Yes, although you come back to the same place, friction has opposed the motionthe entire time.

Name _______________________

Physics 201 practice exam 3 Page 2

3) A glass ball collides head-on with a steel ball (about the same size, butheavier) initially not moving. In the space below, draw the momentumdiagrams for the system before and after the collision.

4) A soapbox car has an initial energy distribution as seen in the first chart. Inthe remaining charts, draw the amount of kinetic energy there must be in thecar and the raceway. (Thermal Energy = Internal energy.)

Energy Diagram

0

10

20

30

40

50

60

Potential Energy Kinetic Energy Thermal Energy

Ener

gy

Energy Diagram

0

10

20

30

40

50

60

Potential Energy Kinetic Energy Thermal Energy

Ener

gy

Energy Diagram

0

10

20

30

40

50

60

Potential Energy Kinetic Energy Thermal Energy

Ener

gy

Energy Diagram

0

10

20

30

40

50

60

Potential Energy Kinetic Energy Thermal Energy

Ener

gy

pglass before psteel before

pglass after psteel after

ptotal

Name _______________________

Physics 201 practice exam 3 Page 3

5) An 80 kg skier travels 400 m down a slope that starts out at a 40o angle andgradually flattens out to a 20o angle, resulting in a total vertical drop of 200meters. The initial push gives her a speed of 3.5 m/s, and she reaches thebottom traveling at 21.5 m/s. What was the average force of friction?

GPEt + KEt = KEb + IEbmgh + ½mvt

2 = ½mvb2 + F(f)d

F(f)*400 m = 80kg*9.8m/s2*200m + ½(80kg)(3.5m/s)2 – ½(80kg)(21.5m/s)2

F(f) = 347 N

6) Tarzan (mass of 75 kg) rescues Jane by swing in at 4.5 m/s and grabbing herso that they then swing off at 2.7 m/s. What must be Jane’s mass?

mtvb = mtva + mjva

75kg * 4.5 m/s = 75kg * 2.7 m/s + mj*2.7 m/s

mj = 75 kg * (4.5 m/s – 2.7 m/s)/2.7 m/s = 50 kg

GPE KE IE GPE KE IE

m = 80 kgh = 200 md = 400 mvt = 3.5 m/svb = 21.5 m/sg = 9.8 m/s2

F(f) = ?

mt = 75 kgmj = ?vb = 4.5 m/sva = 2.7 m/s

Name _______________________

Physics 201 practice exa

7) A 40 g Velcro dart is launched from a spring-loaded gun. It sticks to a fuzzydie (mass 100 g) hanging from the ceiling. The dart and the die swing back,reaching a maximum height of 10 cm above where the die started. How fastwas the dart traveling before it hit the die?

Two parts: first a collision, then a change in height

p1b = p1a + p2a m1v1 = (m1+m2)v2

v2 = sqrt(2gh) = sqrt(2*9

v1 = (m1+m2)v2/m1 = (14

m1 = 40 gm2 = 100 gv1 = ?v2 = ?h = 0.10 m

KE

m 3

KE½(m

.8m/s2 *

0g/40g)*

GPE

= GPE1+m2)v2

2 =

0.1 m) = 1.9

1.96 m/s =

KE

(m1+m2

6 m/s

6.86 m/s

GPE

Page 4

)gh

Name _______________________

Physics 201 practice exam 3

8) Real world problem. Fill in ONLY the indicated steps of the FOCUS andDESCRIBE steps. DO NOT SOLVE THE PROBLEM

As a concerned citizen, you have volunteered to serve on a committee investigating injuries toMiddle School students participating in sports programs. Currently your committee isinvestigating the high incidence of ankle injuries on the basketball team. You are watching theteam practice, looking for activities which can result in large horizontal forces on the ankle.Observing the team practice jump shots gives you an idea, so you try a small calculation. A 40-kg student jumps 1.0 m straight up and shoots the 0.80 kg basketball at his highest point. Fromthe rajectory of the basketball, you deduce that the ball left his hand at 30 from the horizontal at20 m/s. What is his horizontal velocity when he hits the ground?

EVERYDAY LANGUAGE What are you trying to find?Sketch with Given Information

final horizontal velocity

What are the physics principle(s)?momentum

Physics Description Define VariablesDiagram

Quantitative Relationships (Writproblem.)0 = pball after – pkid afterp = mvvball-x = vballcos(θ)

20 m/s @ 30o

pkid before pball before

total momentum

pkid after (horizontal)

pkid before = 0 kgm/spball before = 0 kgm/smkid = 40 kgmball = 0.80 kgpkid after = ?pball after = ?vball = 20 m/svball-x = ?vball-y = ?θ = 30o

pball after (horizontal)

Page 5

e down ONLY the equations needed to solve this

vkid-x = ?