Q A billiard ball (A) with a mass of 0.165 kg
is traveling at 4.2 m/s. It strikes ball B, which is
identical to ball A and is stationary. Ball moves on
at angle of 25 degrees to the left of its original
direction, which ball B moves 65 degrees to the
right. a)Draw a vector diagram to show how the law of
conservation of momentum could be applied to this
situation. b)Write a simplified equation, in terms of
mass and the velocity variables, to show how the law
of conservation of momentum could be applied to this
situation. c)Use the vector diagram and suitable
calculation to determine speeds of ball A and ball B
after the collision.
Ans
Conservation
of momentum will be applied to x and y directions
seperately
Along
x axis
ma
ua = ma va Cos 25 +
mb vb Cos 65
4.2 =
0.91va + 0.42vb ----- 1
Along
y axis
0 = ma
va Sin 25 - mb vb
Sin 65
0.42 va
= 0.91 vb
or va
= 2.17vb
substituting
in eq 1 and solving we get
vb
= 1.75 m/s
hence va
= 2.17 vb = 3.8 m/s
Q A singly ionized particle (q = 1.60 x 10-19
C) is moving at 1.9 x 104 m/s and moves
perpendicular into a magnetic field of 1.0 x 10-3
T. If the particle moves in a radius of curvature of 0.40 m,
determine the mass of the particle. Begin your solution with
a statement about forces.
Ans radius of curvature r = 0.4 m
magnetic field B = 1 x 10-3 T
charge = 1.6 x 10-19 C
velocity v = 1.9 x 104 m/s
A force acts
on the charged particle in a magnetic field which is given by
F= qv x B. Direction of force is given by right hand screw
rule. In this case it is perpendicular to the paper,
outwards. A force perpendicular to the direction of motion
can change only the direction of motion and therefore makes
it move in a circular path. This force thus provides the
centripetal force.
F =qv x B =
qvBSin() = qvB = mv2/r
where m is the
mass of the particle and r is radius of the path
m = Bqr/v = (1
x 10-3 x 1.6 x 10-19 x 0.4) / 1.9 x 104
= 0.34 x 10-26 kg
Q A plane
fires a bullet with an initial velocity of 100m/s at an angle
of 37 degrees below the horizontal. The bullet lands two
seconds later. In the questions that follow, you may neglect
air resistance and assume that g = 10 m/s/s.
a. Find the
horizontal (x) and the vertical (y) components of the
bullet's initial velocity.
b. How far
does the bullet travel in the horizontal direction before it
lands? Show your work.
c. How high
above the (level) ground was the plane when the bullet was
fired? Show your work.
d. Find the
magnitude of the velocity of the bullet right before it hits
the ground. Show your work.
Ans a
u = 100 m/s
angle A = 37 degrees
g = 10 m/s2
t = 2 s
Horizontal component of bullets initial velocity = ux
= u Cos 37 = 80 m/s Vertical component of bullets initial velocity = uy
= u Sin 37 = 60 m/s
Ans b
The
acceleration in the horizontal direction = ax = 0
therefore distance traveled in the horizontal direction x = ux
t
here ux has been found out to be 80 m/s
therefore x = 80 x 2 = 160 m
Ans c
Acceleration
in the vertical direction = g = 10 m/s2
Distance traveled in the vertical direction
h = uyt
+ ½ ayt2 = uyt + ½ gt2
= 60 x 2 x ½
x 10 x 4 = 100 m
Ans d
vx
= x (horizontal) component of final velocity
vy
= y (vertical) component of final velocity
vx
= ux + ax t = 80 m/s as acceleration in
horizontal direction is zero
vy
= uy + gt = 60 + 10x2 = 80 m/s
v = final
velocity =(vx2 + vy2)1/2
= (802 + 802)1/2 =113.14
m/s
Angle of v = B
= tan-1vy/vx = tan-1
1= 45 degrees
Q. Determine
the kinetic energy of an electron that has a wavelength of
1m.
A. = λ h/p
h = 6.63 x 10 -34 Js
or p = h/λ = (6.63 x 10 -34Js)/1m
k = p2/2m = (6.63 x 10 -34)2
/ 2 x 9.1 x 10 -31 = 2.4 x 10 -37 J
= (2.4 x 10 -37) / 1.6 x 10 -19 = 1.5 x
10 -18 eV
Q An observer
looks at a source through a grating having 1000 slits per cm.
If the wavelength is 550nm, how many images of the source can
be seen? (Hint: For what order would theta be + or - 90
degrees)
A.
d = grating
spacing
deviation angle = 30 degs
λ = wave length of light
n = spectral order
N = no of lines / cm = 1000
N= 1/d or d = 1/N = 1/103 cm = 10-3 cm
= 10 -5 m
nλ = d Sin A
λ = 550
nm = 550 x 10 -9 m.
Maximum number of images will be seen for greatest angle of
deviation i.e. A = + or - 90 degs
hence Sin A = + or - 1
therefore the order numbers for the visible spectrum for the
grating are limited to
n < = d / λ = 10 -5 / 550 x 10 -9 =
18.2
hence 18 complete orders will be observed, or 36 images;18 on
each side of central maximum
(x-1)mo c2 = mo c2 {1 / (1 - v2 / c2) - 0.5}
- 1 = 1
1 / (1 - v2 / c2) - 0.5 = 2
1 - v2 / c2 = 1/4
v = 0.866c
Q. Draw a
motion diagram and a free-body diagram for a roller coaster
car at the top of a "loop-de-loop" (i.e., when
the car is upside down). Do not ignore air resistance,
and assume that the car is moving rapidly enough to remain in
firm contact with the rails. In particular, explain why
there is no outward force on the car.
A. The forces
on the roller coaster are ( in the absence of air friction
and friction between rails and the car)
a. Weight mg downwards
b. Normal
force N by the rails on the
car, also downwards
From Newton's
second law N + mg = (mv2 )/ r = centripetal force
to execute the circle with velocity v
Min speed at
the top = (rg)-2 where r is the radius of the loop
This speed is
derived from conservation of energy. If air friction is taken
into account, then v must be greater than (rg)-2
as part of the energy will be used to overcome friction.
Outward force
or cetrifugal force is a pseudo force which needs to be taken
into account if we describe the motion of the car on the
roller coaster from a rotating frame which is non inertial
and still use Newton's laws. As we are observing the roller
coaster from the ground frame, which is inertial, there is no
pseudo force.
Q. A coil of
diameter 3 m having 120 turns and resistance 8.0 m ohms is
flat on a table, and a magnetic induction perpendicular to
its plane changes at a steady rate from 0 to 0.8T. How much
charge flowed through the coil during the time required for
the field to change?
A Flux Ø =
nAB
dØ = -nAdB
dØ/dt =
-nAdB/dt = -e
this implies
that e = nAdB/dt =iR
which implies
that nAdB = iRdt
nAdB = Rdq
Integrating
both sides of the above equation with limits
q = nAB/R =
nAx0.8/R
= (120 x pi x
(1.5)2 x 0.8)/8 = 84.78 Coulomb
Q The ac
generator in Fig below supplies 120 V (rms) at 60 Hz. With
the switch open as in the diagram, the resulting current
leads the generator emf by 20 degs. With the switch in
position 1 the current lags the generator emf by 10 deg. When
the switch is in position 2 the rms current is 2.0 A. find
the values of R,L, and c.
Ev
= 120V , f=60 Hz , w=2pif = 360 radians
Current I
leads by 20 degs
Phase is
negative, Xl < Xc
Tan (-20deg)=
(Xl - Xc)/R = -0.36 ---- (1)
Or (Xc
-Xl)/R =0.36
I lags by 10
degs and phase is positive
Xl
> Xc1
Tan 10 degs =
0.18 = (Xl - Xc1)/R ------(2)
Z=Xc
- Xl = Ev /Iv = 120/2 =60
---------(3)
Substituting 3
in 1
60/R=0.36
R=60/0.36 =
500/3 ohms
Equation 2
implies wL-1/2wC = 0.06 x 500/3 = 30
Equation 3
implies 1/wC -wL=60 ----------(a)
wL = 30 +
1/2wC ------(b)
substituting
in a
1/wC -30 -
1/2wC =60
1/2wC = 90
-------------(c)
C=0.15 µF
Substituting c
in b
wL=30 + 90
L= 1/3 H
Q Find the
magnetic field at the center of the loop shown.
B = (µo/4pi)(IL/r2)
B = 0
Since I1R1
= I2R2 or I1L1 =
I2L2
Or B = (µo/4pir2)(
I1L1 - I2L2) = 0
Q What is the
back emf in a motor that draws 7.1 A from a 220 volt line if
the field coils have a resistance of 200 ohms and are wound
in parallel with an armature with a resistance of 3 ohms?
I=7.1 A , v =
220 V , R1 = 200 ohms, R2 = 3 ohms
Let E = back
emf
I =(V-E)/R
1/R =1/3
+1/200 ie R = 600/203
E = V - IR =
220 - 21 = 199 V
Q. A
horizontal rod 2 meters long is released in a region where
there is a uniform horizontal magnetic induction of 0.6 Tesla
perpendicular to the rod. How far has the rod fallen when the
induced emf is 120V.
