PHYS101 Interm Exam - Solution SetDepartment of Physics

Summer 2014 - September 08, 2014

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Questions:

1. A 1200 kg car has a horizontal speed v = 18.4m/s, when it hits a horizontal springand comes to rest in a distance of 2.3 m. Find the spring constant. (4P)

Solution:

The car-spring system is isolated without friction =⇒ ∆Emech = 0

(KC f + Usp f

)−(KCi + Uspi

)= 0(

0 +12

kx2)−(

12

mv2i + 0

)= 0

12

kx2 =12

mv2i

k = mv2

ix2 = 1200kg

(18.4m

s2.3m

)2

= 76800Nm

= 76.8kNm

1

2. A person pulls from rest a 60 kg block 50 m along a horizontal rough surface with akinetic frictional constant of µk = 0.05 by a constant force Fapp = 120N at an angleθ = 42◦ with the horizontal, as shown in the figure below. (6P)

(a) What is the work done by the force ~Fapp on the block?

(b) What is the work done by the kinetic frictional force?

(c) What is the final velocity of the block after beingpulled for 50 m?

Solution:

(a) Work done by the applied force:

Wapp = ~Fapp · (∆x i) = Fapp cos θ ∆x = 120N cos (42◦) 50m = 4458.9J

(b) Work done by the kinetic frictional force:First we have to calculate the kinetic frictional force. Obviously FN = Fg −Fapp sin θ, therefore we get for the kinetic frictional force

fk = FNµk =(

Fg − Fapp sin θ)

µk =(mg− Fapp sin θ

)µk =

=(

60kg 9.80ms2 − 120N sin 42◦

)0.05 = 25.4N.

Then we can calculate the work done by the kinetic frictional force

W f r = − fk∆x = −29.4N 50m = −1270J

(c) The final velocity of the box:Using the Work-Kinetic Energy Theorem:

∆K = ∑ Wext

12

mv2f −

12

mv2i = Wapp + W f r

12

mv2f − 0 = Wapp + W f r

v f =

√2m(Wapp + W f r

)=

√2

60kg(4458.9J − 1270J) = 10.3

ms

2

3. A puck with mass m2 = 0.3kg , initially at rest on a horizontal, frictionless surface, isstruck by a puck of mass m1 = 0.2kg moving initially along the x axis with a speedof v1i = 2.0m/si. After the collision, the puck with mass m1 = 0.2kg has a speed ofv1 f = 1.0m/s at an angle of θ = 53.0◦ to the positive x axis (see figure below). (7P)

(a) Determine the magnitude v1 f and direction φ ofthe velocity of the puck of mass m1 = 0.3kg afterthe collision.

(b) Find the magnitude of kinetic energy transferredaway or transformed to other forms of energy inthe collision.

Solution:

(a) Inelastic Collision, therefore only conservation of momentum

m1~v1i + m2~v2i = m1~v1 f + m2~v2 f

m1v1i i + 0 = m1v1 f (cos θ i + sin θ j) + m2v2 f (cos φi + sin φj)

in component form

m1v1i = m1v1 f cos θ + m2v2 f cos φ (1)

0 = m1v1 f sin θ + m2v2 f sin φ (2)

From (1) and (3) we get:

m2v2 f cos φ = m1v1i −m1v1 f cos θ (3)

m2v2 f sin φ = −m1v1 f sin θ (4)

Dividing equation (4) by (3) gives:

tan φ =−m1v1 f sin θ

m1v1i −m1v1 f cos θ=−v1 f sin θ

v1i − v1 f cos θ

then we get for θ:

φ = tan−1

(−v1 f sin θ

v1i − v1 f cos θ

)= tan−1

( −1.0ms sin 53◦

2.0ms − 1.0 m

s cos 53◦

)= −29.7◦

From (4) we get:

v2 f = −m1v1 f sin θ

m2 sin φ= −

0.2kg 1.0 ms sin 53◦

0.3kg sin(−29.7◦)= 1.07

ms

(b) Energy difference

|∆K| =∣∣∣∣(1

2m1v1 f +

12

m2v2 f

)−(

12

m1v1i +12

m2v2i

)∣∣∣∣ ==

∣∣∣∣(12

0.2kg(

1.0ms

)2+

12

0.3kg(

1.07ms

)2)−(

12

0.2kg(

2.0ms

)2+ 0)∣∣∣∣ = 0.13J

3

4. Consider two objects with m1 = 2kg and m2 = 1kg connected by a light string thatpasses over a pulley having the moment of inertia I = 0.5kgm2 and the radius R =0.3m about the axis of rotation as shown in the figure below. The string does notslip on the pulley or stretch. The pulley turns without friction. The two objects arereleased from rest separated by a vertical distance 2h, with h = 1.5m. (6P)

R

(a) Use the principle of conservation of energy to find thetranslational speeds of the object as they pass eachother.

(b) Find the angular speed at this time.

Solution:

(a) The system of mass m1, mass m2, pulley, earth is isolated without friction. There-fore the mechanical energy of the system is conserved.

∆Emech = 0(K1 f + U1 f + K2 f + U2 f + Krot f

)−(K1i + U1i + K2i + U2i + Kroti

)= 0(

12

m1v2 + m1gh +12

m2v2 + m2gh +12

Iω2)− (0 + m1g(2h) + 0 + 0 + 0) = 0

with ω = vR we get:(

12

m1v2 + m1gh +12

m2v2 + m2gh +12

I( v

R

)2)− (0 + m1g(2h) + 0 + 0 + 0) = 0

Now we have to solve the equation for v:

12

(m1 + m2 +

IR2

)v2 = (2m1 −m1 −m2)gh

v =

√2(m1 −m2)ghm1 + m2 +

IR2

v =

√√√√2(2.0kg− 1.0kg)9.80 ms2 1.5m

2.0kg + 1.0kg + 0.5kgm2

(0.3m)2

= 1.85ms

(b)

ω =vR

=1.85m

s0.3m

= 6.18rad

s

4