EXAM 3 SOLUTIONS - PSL Main Page | Plasma …psl.physics.auburn.edu/etjr/class_f14/pdfs/exam3_f14...... A 300 kg (including passengers) roller coaster car is at the top of a 12 m radius

Exam 3 – Version 1 1

EXAM 3

SOLUTIONS

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Instructions: Read each question CAREFULLY and answer all parts. Work MUST be shown to receive credit. ---------------------------------- MULTIPLE CHOICE ANSWERS HERE ------------------------------- Multiple choice V1 V2

1 d b 2 b e 3 a c 4 c b 5 b a


Part 1: Problems 1 through 5 (6 pts each – 30 pts. total) Multiple Choice Problems Circle your answer AND write your choice (a-e) on the front page 1) A car turns a corner on a banked road near its maximum speed. Which of the diagrams is the car’s free-body diagram?

a) b) c) d) e)

V1 V2 max min d b

2) A 500 kg car goes around a curve at a constant speed of 10 m/s and experiences a net force of 1250 N. The radius of the curve is: a) 0.15 m b) 40.0 m c) 4.0 m d) 0.025 m e) 25.0 m 3) A 12 kg block is moved with a force F = 8 i - 3 j [N] through a displacement of ∆r = 1 i + 2 j [m]; the work done by this force is: a) 2 J b) 6 J c) - 4 J d) -2 J e) 8 J

V1 V2

M [kg] 500 800 V [m/s] 10 10 F [N] 1250 2500 R = mv2/F [m] 40 32

V1 V2

Fx 8 -‐6 Fy -‐3 4 rx 1 1 ry 2 3 W = F•r = Fxrx + Fyry 2 J 6 J


4) A 0.05 kg tennis ball is travelling straight at a player at 13 m/s. The player hits the ball straight back at 20 m/s. If the ball remains in contact with the racket for 0.082 s, determine the magnitude of the average force that acts on the ball. a) 4.3 N b) 402 N c) 20.1 N d) 70.3 N e) 1.6 N 5) Ball 1 has 3 times the mass of Ball 2. Both are thrown so that they have the same kinetic energy. The velocity of Ball 1 is V. The velocity of Ball 2 is: a) 3.0 V b) 1.7 V c) 9.0 V d) 0.57 V e) 6.2 V

V1 V2

M [kg] 0.05 0.08 Vi [m/s] 13 12 Vf [m/s] -‐20 -‐20 ∆t [s] 0.082 0.082 F = [M*(Vf-‐Vi)]/∆t [N] -‐20.1 -‐31.2

V1 V2

m1 3 5 m2 1 1 V1 1 1 V2 = SQRT(m1/m2)*V1 1.73 2.24 Solution equation based on: ½ m1 v1

2 = ½ m2 v22


[For all of the remaining problems, show your algebraic steps towards a solutions.] Problems 6 to 8: Short Answer Problems: (10 pts each) 6) A 4 kg toy is pushed across a floor with a force given by the plot below with F0 = 30 N. If the car starts at rest, use the work-energy theorem to find the velocity of the car after it has moved 0.2 m.

7) A 300 kg (including passengers) roller coaster car is at the top of a 12 m radius loop. Find the minimum speed the coaster car must have in order to remain on the track.

8) A box of mass M and velocity V (+i) makes a perfectly inelastic collision with a box of mass 3M that has a velocity 0.5V (+i). What is the velocity of the system after the collision.

V1: Conserve momentum: MV + 3M(0.5V) = (M+3M)Vf

Vf = MV +3M(0.5V ) (M +3M)

= 2.5MV4M

=58

V = 0.625V

V2: Conserve momentum: MV + 2M(0.75V) = (M+2M)Vf

Vf = MV +2M(0.75V ) (M +2M)

= 2.5MV3M

=56

V = 0.833V

r"="0.2"m"

F0"

r"

m 3m#

W = ∆K = ½ mvf2 – ½ mvi

2 W = Area under curve of F(x) = ½ F0 r Vf = [F0r/m]1/2

V1 V2

F0 [N] 30 40 r [m] 0.2 0.2 m [kg] 4 3 v [m/s] 1.22 1.63

V1 V2

m [kg] 300 400 r [m] 12 8 v [m/s] 10.84 8.85


Extended Problems: These problems at 20 points each. You need to choose 2 out of the 3 problems. Indicate which two problems you want to have graded using the box AND on the front of the exam. Scoring: invalid or no units [-1/each]; no calculus/algebraic work shown [-3]

Extended Problem 1: [Parabolic trajectory] A child of mass m starts at rest and slides without friction from a height h = 3.6 m along a water slide that is next to a pool. She is launched from a height h/5 into the air over the pool at an angle of θ = 25˚. Solve parts (a) and (b) using energy techniques. a) Find the girl’s velocity vector the instant she leaves the slide. [7 pts] b) Find the maximum height of the girl while she is in the air. [7 pts] c) Find the distance the girl lands from the end of the slide. Solve using any appropriate method [6 pts]

Part (a):Ui +Ki =Uf +Kf

mgh+0 =mg h5

!

"#$

%&+

12

mv 2 ⇒ v 2 = 2gh h− h5

!

"#

$

%&⇒ v = 8

5gh

This is the magnitude of the velocity, the question asks for velocity vector:!v = v cosθ i +v sinθ j

Part (b):

Start from slide where: v i2 = vx

2 +vy2( ) [from Part (a)]

Ui +Ki =mg h5

!

"#$

%&+

12

mv 2 =mg h5

!

"#$

%&+

12

m vx2 +vy

2( )At the maximum height, the vertical velocity is zero, so v f

2 = vx2( )

Uf +Kf =mgymax +12

m vx2( )

Solve for ymax from conservation of energy: ymax =h5+

vy2

2g


Part (c): Best to use kinematics: solve for time in the air, then solve for distance

You have: yi =h5

,yf = 0,vyi = v sinθ,vxi = v cosθ,xi = 0,ax = 0,ay = −g

Use : yf = yi +vyit +12

ayt2 to solve quadratic equation for time: t =

−vyi + vyi2 −4 h

5

#

$%&

'( −

g2

#

$%

&

'(

2 −g2

#

$%

&

'(

Then :d = vxit

V1 V2

h [m] 3.6 4.2 h/5 [m] 0.72 0.84 Ang (deg.) 25 22 v [m/s] 7.51 8.12 vx [m/s] 6.81 7.52 vy [m/s] 3.18 3.04 ymax [m] 1.23 1.31 t [sec] 0.18 0.21 d = vx*t [m] 1.21 1.56


Extended Problem 2: [Spring-mass system] A box of mass m = 3.0 kg is compressed a distance x against a spring with k = 2500 N/m. When released, the mass travels up an inclined plane as shown in the figure. At the equilibrium length of the spring, the mass leaves the spring and travels, with friction, along the length of inclined plane a distance, d = 0.6 m before coming to rest. Let the coefficient of kinetic friction be µk = 0.3 and the angle of the inclined plane be θ = 37˚. Find the distance x that the spring must be compressed. The horizontal dashed line is the point at which the mass leaves the spring. Solve using energy techniques. [potential energy terms – 9, dissipation term – 3, algebraic setup – 4, solution – 4]

Use horizontal dashed line as reference for potential energy for spring and gravity (y = 0)

Conservation of Energy:Uf +Kf + ∆Eint =Ui +Ki where: ∆Eint = -Wres is the change in internal energy (ie, due to friction) = fkd

where : fk = µkN = kinetic friction

Block starts and ends at rest: Ki =Kf = 0

Ui =Usp−i +Ugrav−i =12

kx2 +mg(−xsinθ)

Uf =Ugrav−f =mg(d sinθ)

∆Eint = −Wres = fkd = µkN = µkmgcosθ

Resulting equation for conservation of energy is a quadratic equation in "x":12

kx2 +mg(−xsinθ) =mgd sinθ + µkmgd cosθ

12

k#

$%

&

'(x2 − (mgsinθ)x −mgd(sinθ + µk cosθ) = 0

Solve the quadratic equation for “x”.

m"

d"

x"

θ


V1 V2

m [kg] 3 2 k [N/m] 2500 3500 d [m] 0.6 0.8 ang (th) 37 37 µk 0.3 0.3 Ff=µk*mgcos(th) [N] 7.04 4.70 h = d*sin(th) [m] 0.36 0.48 Ff*d [J] 4.23 3.76 mgh = mgdsin(th) [J] 10.62 9.44

A = 0.5k 1250 1750 B = -‐mgsin(th) 17.69 11.80 C = -‐(Ff*d+mgh) -‐14.84 -‐13.19

x+ [m] 0.102 0.084


Extended Problem 3: [Collision] A 2.25 kg wooden block of mass M rests on a table over a large hole. A 10 gram bullet is fired with an initial speed vi is fired upward into the bottom of the block. The bullet passes through the block and emerges with a speed 0.2vi. As a result of impact, the block rises into the air to a maximum height of 18 cm. Assume the bullet makes a perfect hole through the block (i.e., the block does not lose mass) and that “upward” is the +y-direction. a) Find the initial speed of the bullet. [10 pts] b) Find the impulse vector on the bullet. [5 pts] c) The bullet took ∆t = 2 ms to pass through the block, find the average force vector on the block. [5 pts] Two step problem: solve in reverse order – first is the conservation of energy, then conservation of momentum. Part (a):

Conservation of Energy (for the Block):Ki +Ui =Kf +Uf ; where: Ui → 0 [block is at y = 0] and K f → 0 [at max height, vy = 0]

So :12

MBV2 =MBgymax →V = 2gymax - velocity of the Block after the collision

Conservation of momentum: (b = bullet, B = block)mbvbi +MBvBi =mbvbf +MBvBf →mbvbi +0 =mbvbf +MBV

So : (vbi −vbf ) =MB

mb

V;

V1: 0.8v i =MB

mb

V

V 2 : 0.7v i =MB

mb

V

Part (b):

!J = ∆

!p =m

!vf −!vi( )

Since the final velocity is lower than the initial, the impulse vector will be in -y (-j) direction


V1 V2

MB [kg] 2.25 4.5 coeff 0.2 0.3 mb [kg] 0.01 0.01 h [m] 0.18 0.12 ∆t [seconds] 0.002 0.002 V [m/s] 1.88 1.53 vi [m/s] 528 986 J (-‐y) [N-‐s] 4.23 6.90 F (+y) [N] 2113 3451

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EXAM 3 SOLUTIONS - PSL Main Page | Plasma …psl.physics.auburn.edu/etjr/class_f14/pdfs/exam3_f14...... A 300 kg (including passengers) roller coaster car is at the top of a 12 m radius