PHYS101 Midterm Exam Solution-Set
Department of Physics
Spring 2014 - April 11, 2014
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c
2014
Department of Physics, Eastern Mediterranean University
Questions:
1. A small sphere of mass m = 1 kg is attached to the end of a massless cord of length
R = 1 m and set into motion in a horizontal circle about a fixed point O on a frictionless surface as shown in the figure below. (The cord remains horizontal during the
motion)
horizontal xy-plane
(a) Draw the free body diagram for the mass m.
O
Solution:
(a)
R
m
(b) If the tension of the cord is T = 16N, calculate
the speed of the sphere.
(b)
O
v2
∑ Fx = −T = −mar =⇒ T = m r =⇒
s
r
Tr
16N · 1m
m
v =
=
=4
m
1kg
s
1
2. Two objects are connected by a light string that passes over a frictionless and massless
pulley as shown in the figure below. Assume the incline is a rough surface with
friction and take m1 = 1 kg and the angle is θ = 37◦ . The mass m2 is just about to
slide down the incline. (Objects are not moving)
(a) Draw the free body diagrams for the masses m1 and m2 .
Solution:
(b) For the given coefficient of static friction µs = 0.5 calculate the maximum mass
m2 .
Solution:
From the Free Body diagram of m1 we get:
∑ Fy = T − m1 g = 0 =⇒ T = m1 g
From the Free Body diagram of m2 we get:
∑ Fx
∑ Fy
= Fg2 sin θ − T − f s = 0
(1)
= FN − Fg2 cos θ = 0
(2)
From (2) we get
FN = m2 g cos θ =⇒ f s = µs FN = µs m2 g cos θ
(3)
(3) in (2) gives
m1
sin θ − µs cos θ
1kg
m2 =
= 4.94kg
◦
sin 37 − 0.5 cos 37◦
m2 g sin θ − m1 g − m2 g cos θµs = 0 =⇒ m2 =
2
3. A particle moves along the x-axis. It’s position is given by the equation
x (t) = 2 + 3t − t2 , with x (t) in meters and t in seconds.
(a) Find the magnitude of the displacement the particle undergoes between t = 1s
and t = 3s.
Solution:
x (1s) = 2 + 3 − 1 = 4m, x (3s) = 2 + 9 − 9 = 2m
Then we get for the magnitude of Displacement |∆x | = | x (3s) − x (1s)| = 2m.
(b) Find the magnitude of the average velocity in the time interval t = 1s and t = 3s.
Solution:
Magnitude of the average velocity in the time interval ∆t = 2s is given as
|v avg | =
2m
m
|∆x |
=
=1
|∆t|
2s
s
(c) Determine the position of the particle when it changes its direction.
Solution:
The point where the particle changes is direction is determined by the root of the
velocity, therefore we have to determine first the instantaneous velocity v(t).
v(t) = x 0 (t) = 3 − 2t
(4)
v(t) = 0 corresponds tot t = 3/2, so we get for the position where this happens
x (1.5s) = 4.25m.
(d) Determine the particles velocity when it returns to the position it had at t = 0.
Solution:
First we have to calculate the time the particle needs to return to its starting
position at t = 0.
x (t) = 2 + 3t − t2 = x (0) =⇒ 3t − t2 = 0 ⇐⇒ (3 − t)t = 0 ⇐⇒ t = 0 or t = 3s
As t = 0 is the starting point, we need to calculate the velocity (4) at t = 3s
v(3s) = 3 − 2 · 3 = −3m/s
4
3
2
1
1
2
3
-1
-2
3
4
4. A movie stunt driver on a motorcycle speeds horizontally off a 50m high cliff. If the
motorcycle will land 90m from the base of the cliff, (ignore any kind of friction or
resistance)
y
(a) Find the time of flight,
(b) Find its initial speed in x-direction v x0 ,
(c) Find its acceleration vector just before
hitting the ground.
O
x
(a) Solution:
As in this case the height of the cliff is given and the y-component of the initial
velocity is 0, then we can calculate the time of flight as follows:
s
s
1 2
2y0
2 · 50m
1 2
=
y(t) = y0 + vy0 t − gt = y0 − gt = 0 ⇒ t =
= 3.19s
2
2
g
9.8 sm2
(b) Solution:
As we have the time of flight, and the range of the projectile motion, keeping
in mind that there is no acceleration in x-direction, we can calculate the initial
velocity as follows
x (t) = x0 + v x0 t = v x0 t =⇒ v x0 =
90m
m
x (t)
=
= 28.21 .
t
3.19s
s
(c) Solution:
The acceleration of the particle is at every point of its motion
~a = − gĵ.
4