PHYS101 Interm Exam - Solution Set

PHYS101 Interm Exam - Solution Set
Department of Physics
Summer 2014 - September 08, 2014
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c
2014
Department of Physics, Eastern Mediterranean University
Questions:
1. A 1200 kg car has a horizontal speed v = 18.4m/s, when it hits a horizontal spring
and 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
1 2
1 2
mv + 0
= 0
0 + kx −
2
2 i
1 2
1 2
kx =
mv
2
2 i
18.4 ms 2
v2i
N
kN
k = m 2 = 1200kg
= 76800 = 76.8
2.3m
m
m
x
1
2. A person pulls from rest a 60 kg block 50 m along a horizontal rough surface with a
kinetic frictional constant of µk = 0.05 by a constant force Fapp = 120N at an angle
θ = 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 being
pulled for 50 m?
Solution:
(a) Work done by the applied force:
Wapp = ~Fapp · (∆x ˆi) = 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
f k = FN µk = Fg − Fapp sin θ µk = mg − Fapp sin θ µk =
m
◦
= 60kg 9.80 2 − 120N sin 42 0.05 = 25.4N.
s
Then we can calculate the work done by the kinetic frictional force
W f r = − f k ∆x = −29.4N 50m = −1270J
(c) The final velocity of the box:
Using the Work-Kinetic Energy Theorem:
∆K =
∑ Wext
1 2 1 2
mv f − mvi = Wapp + W f r
2
2
1 2
mv f − 0 = Wapp + W f r
2
s
r
2
2
m
vf =
Wapp + W f r =
(4458.9J − 1270J ) = 10.3
m
60kg
s
2
3. A puck with mass m2 = 0.3kg , initially at rest on a horizontal, frictionless surface, is
struck by a puck of mass m1 = 0.2kg moving initially along the x axis with a speed
of v1i = 2.0m/sˆi. After the collision, the puck with mass m1 = 0.2kg has a speed of
v1 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 of
the velocity of the puck of mass m1 = 0.3kg after
the collision.
(b) Find the magnitude of kinetic energy transferred
away or transformed to other forms of energy in
the collision.
Solution:
(a) Inelastic Collision, therefore only conservation of momentum
m1~v1i + m2~v2i = m1~v1 f + m2~v2 f
m1 v1i ˆi + 0 = m1 v1 f (cos θ ˆi + sin θ ˆj) + m2 v2 f (cos φˆi + sin φˆj)
in component form
m1 v1i = m1 v1 f cos θ + m2 v2 f cos φ
(1)
0 = m1 v1 f sin θ + m2 v2 f sin φ
(2)
m2 v2 f cos φ = m1 v1i − m1 v1 f cos θ
(3)
m2 v2 f sin φ = −m1 v1 f sin θ
(4)
From (1) and (3) we get:
Dividing equation (4) by (3) gives:
tan φ =
−m1 v1 f sin θ
m1 v1i − m1 v1 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
−1.0 ms sin 53◦
2.0 ms − 1.0 ms cos 53◦
= −29.7◦
From (4) we get:
v2 f = −
m1 v1 f sin θ
m2 sin φ
0.2kg 1.0 ms sin 53◦
m
=−
= 1.07
◦
0.3kg sin(−29.7 )
s
(b) Energy difference
1
1
1
1
m1 v1i + m2 v2i =
|∆K | = m1 v1 f + m2 v2 f −
2
2
2
2
m 2 1
m 2
2
1
m
1
= 0.2kg 1.0
+ 0.3kg 1.07
−
0.2kg 2.0
+ 0 = 0.13J
2
s
2
s
2
s
3
4. Consider two objects with m1 = 2kg and m2 = 1kg connected by a light string that
passes 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 not
slip on the pulley or stretch. The pulley turns without friction. The two objects are
released from rest separated by a vertical distance 2h, with h = 1.5m. (6P)
(a) Use the principle of conservation of energy to find the
translational speeds of the object as they pass each
other.
R
(b) Find the angular speed at this time.
Solution:
(a) The system of mass m1 , mass m2 , pulley, earth is isolated without friction. Therefore 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
1
1 2
1
2
2
m1 v + m1 gh + m2 v + m2 gh + Iω − (0 + m1 g(2h) + 0 + 0 + 0) = 0
2
2
2
with ω = Rv we get:
1
1 v 2
1
2
2
m1 v + m1 gh + m2 v + m2 gh + I
− (0 + m1 g(2h) + 0 + 0 + 0) = 0
2
2
2 R
Now we have to solve the equation for v:
1
I
m1 + m2 + 2 v2 = (2m1 − m1 − m2 ) gh
2
R
s
2(m1 − m2 ) gh
v =
m1 + m2 + RI2
v
u 2(2.0kg − 1.0kg)9.80 m 1.5m
m
u
s2
= 1.85
v = t
0.5kgm2
s
2.0kg + 1.0kg + (0.3m)2
(b)
ω=
1.85 ms
v
rad
=
= 6.18
R
0.3m
s
4

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