Free Physics and Chemistry Video Lectures, Theory Notes, Solved Numericals, M.C.Qs For Class IX, Class X, Class XI & Class XII
November 30, 2012
November 29, 2012
November 28, 2012
November 27, 2012
Projectile Motion : Practice Problems + Solutions
Solve the following questions using what you know about
projectile motion.
1. A
roadrunner runs directly off a cliff with an initial velocity of 3.5 m/s.
a) What
are the components of this velocity?
Vx = 3.5 m/s Vy = 0 m/s
b) What
will be the horizontal velocity 2 seconds after the bird leaves the cliff?
3.5 m/s – horizontal velocity is unchanging
c) If
the cliff is 300 m high, at what time will the roadrunner reach the ground?
h = dy = ½ * 10 * t2 =
300
300 * 2 / 10 = t2 = 60
t = 7.75 s
d) How
far from the cliff will this bird land?
dx = 3.5 * 7.75 = 27.125 m
e) If
there is a small pond which begins 25m away from the cliff and extends 2.5
meters from there; will the roadrunner land in the pond?
Yes, the pond is
from 25 m to 27.5 m, so the roadrunner will land in the pond.
f) What
is the final vertical velocity at which the roadrunner is traveling? [The
vertical velocity at the time when the bird reaches the ground]
Vy = 10
* 7.75 + 0 = 77.5 m/s
g) What
is the final horizontal velocity at which the roadrunner is traveling? [The
horizontal velocity at the time when the bird reaches the ground]
Vx = 0
+ 3.5 = 3.5 m/s
h) What
is the total final velocity of this motion? [magnitude and direction]
V2 = 77.52 + 3.52
= 6018.5
V = 77.579 m/s
q = tan-1 (77.5 / 3.5) = 87.41o below
the horizontal
2. An
object (any object) is dropped from a height of 300m
a) How
long does it take this object to fall to the ground?
h = dy = ½ * 10 * t2 =
300
300 * 2 / 10 = t2 = 60
t = 7.75 s
b) Compare
this answer with you answer from question 1, part c). What are the reasons for
any similarities or differences?
They are the same. This is because their
vertical motions are identical. All objects fall with the same gravitational
acceleration, so two objects at the same height with the same initial vertical
velocity will reach the ground at the same time.
3. The
intent of a bean bag toss game is to get your bean bag to land on the
‘bull’s-eye’ of a target. The target is set up parallel to the ground and is
the same height above the ground as your hand is when you let go of the bean
bag. The game’s rules further require you to be 5 m from the center of the
target when you release the bag.
a) Evaluate
the following questions for both an angle of 32o and an angle of 58o
if the bean bag is thrown with an initial velocity of 6 m/s:
32o 58o
i.
What are the components of velocity?
Vx32:
Cos 32o = Vx32/6 Vx58: Cos 58o = Vx58/6
6 * Cos 32o
= Vx32 = 5.09 m/s 6
* Cos 58o = Vx58 = 3.18 m/s
Vy32:
Sin 32o = Vy32/6 Vy58:
Sin 58o = Vy58/6
6 * Sin 32o
= Vy32 = 3.18 m/s 6
* Sin 58o = Vy58 = 5.09 m/s
ii. What
is the maximum height of the bean bag’s motion?
tTOP32
= Vy/10 = 3.18/10 = 0.318s tTOP58
= Vy58/10 = 5.09/10 = 0.509s
hMAX32
= ½ * 10 * 0.3182 = 0.506 m hMAX58
= ½ * 10 * 0.5092 = 1.295 m
iii. How
long will the bean bag be in the air?
tTOTAL32
= 2 * tTOP32 = 0.636 s tTOTAL58
= 2 * tTOP58 = 1.018 s
iv. How
far away from you will the bag land?
dx32
= 5.09 * 0.636 = 3.24 m dx58 = 3.18
* 1.018 = 3.24m
v. If
the center of the bull’s-eye ranges from 4.9 m to 5.1 m away from you, does
your bean bag win?
No No
4. A
stunt driver drives a red mustang convertible up a ramp and off a cliff. The
car leaves the ramp at a velocity of 60 m/s at an angle of 45o to
the horizontal; the cliff and ramp combined cause the car to begin its
projectile motion at a height of 315m above the ground. If you were
coordinating this stunt, how far away would you put a landing surface so that
your stunt driver was not injured?
First let’s think about strategy. The question
is basically asking how far away from the cliff the car will land. In order to
find horizontal distance we need horizontal velocity and time. We can find both
horizontal and vertical velocity from the initial conditions, but we’ll have to
calculate the time it will take for the car to reach the ground. So first we’ll
find the components of velocity:
Vx:
Cos 45o = Vx/60 Vy:
Sin 45o = Vy/60
60 * Cos 45o
= Vx = 42.43 m/s 60
* Sin 45o = Vy = 42.43 m/s
Note: Remember that horizontal velocity is constant, but the
vertical velocity we calculated above is only the initial vertical velocity.
From here we need to use the initial vertical
velocity to find the time it takes the car to reach the top of its path and
fall to the ground. Let’s think about this in two parts; the time it takes to
reach hMAX first:
tTOP = 42.43/10 = 4.243 s
Now what about the time it takes to fall from
the maximum height? Well first we need to know the maximum height:
hTOP = ½ * 10 * 4.2432 =
90.02 m
hMAX = hTOP + ho
= 90.02 + 315 = 405.02 m
Now we calculate the time it takes to fall
from a height of 405.02 m:
405.02 = ½ * 10 * tDOWN2
tDOWN2 = 81.004
tDOWN = 9.000 s
Putting these two times together, we have the
total time it takes the car to travel up to its maximum height and then fall
back down. This is the total time in the air and this is the time we will want
to use to solve for horizontal distance.
tTOTAL = tTOP + tDOWN
= 4.243 + 9.000 = 13.243 s
dx = 42.43 * 13.243 = 561.9 m
You will want to make sure that the landing
surface is centered 561.9 m from the base of the cliff.
Chapter 6: Work and Energy
Practice Problems
Just as a car tops a 38 meter high hill with a speed of 80 km/h it runs out of gas and coasts from there, without friction or drag. How high, to the nearest meter, will the car coast up the next hill?
A pendulum has a mass of 3.6 kg, a length of 1.7 meters, and swings through a (half)arc of 29.4 degrees. What is its amplitude to the nearest centimeter?
To the nearest tenth of a Joule, what is its maximum kinetic energy of the pendulum in problem 2?
To the nearest tenth of a Joule, what is the total energy of the pendulum in problem 2?
A 2 kg metal plate slides down a 13-meter high slope. At the bottom its speed is 9 m/s. To the nearest Joule, what was the magnitude of the work done by friction?
If the slope in the above problem is 23 degrees, what is the coefficent of friction (to 2 decimal places)?
An unstretched spring with spring constant 36 N/cm is suspended from the ceiling. A 3.6 kg mass is attached to the spring and let fall. To the nearest tenth of a centimeter, how far does it stretch the spring?
If the mass in the previous problem is attached to the spring and slowly let down, to the nearest tenth of a centimeter, how far does it stretch the spring?
A mass of 2.2 kg is dropped from a height of 4.7 meters above a vertical spring anchored at its lower end to the floor. If the spring constant is 33 N/cm, how far, to the nearest tenth of a cm, is the spring compressed?
If the top of the spring in the preceding problem is 1.34 meters above the ground when the mass is released, what is the ball's kinetic energy, to the nearest Joule, just before the mass strikes the spring?
Problem 1 The correct answer is 63.
Problem 2 The correct answer is 87.
Problem 3 The correct answer is 7.7.
Problem 4 The correct answer is 7.7.
Problem 5 The correct answer is 174.
Problem 6 The correct answer is 0.29.
Problem 7 The correct answer is 2.
Problem 8 The correct answer is 1.
Problem 9 The correct answer is 25.4.
Problem 10 The correct answer is 101.
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
Answers
Problem 2 The correct answer is 87.
Problem 3 The correct answer is 7.7.
Problem 4 The correct answer is 7.7.
Problem 5 The correct answer is 174.
Problem 6 The correct answer is 0.29.
Problem 7 The correct answer is 2.
Problem 8 The correct answer is 1.
Problem 9 The correct answer is 25.4.
Problem 10 The correct answer is 101.
November 13, 2012
Chapter #10 :Geometrical Optics : Short Q/A / C.R.Q's
Q.Why do thick lenses
possess chromatic and spherical aberrations? Suggest remedies for the
rectification of these defects.
Ans.Chromatic
aberration occurs because lenses have a different refractive index for
different wavelengths of light (the dispersion of the lens) and spherical aberration
is an optical effect observed in an optical device that occurs due to the
increased refraction of light rays when they strike a lens or a reflection of
light rays when they strike a mirror near its edge, in comparison with those
that strike nearer the centre.
Chromatic
aberration was reduced by increasing the focal length of the lens where
possible.
Q. Does the chromatic
aberration takes plane in mirror?
Ans. In optics, chromatic aberration (chromatic distortion) is a type of
distortion in which there is a failure of a lens to focus all colors to the
same convergence point. It occurs because lenses have a different refractive
index for different wavelengths of light (the dispersion of the lens).But A/c
to law of reflection” The angle of incidence is always equal to law of
refraction”. So there is no chance of chromatic aberration in mirror
Chapter #9 :Nature of Light ( Physical Optics) : Short Q/A / C.R.Q's
Q.Why the central
spot in the Newton’s
rings is dark?
Ans. The Newton’s
rings are formed due to the phenomenon of thin film interference. here, the
condition for constructive interference(the ring appearing bright) is that the
optical path difference between interfering waves should be an integral
multiple of the wavelength. the optical path difference is given by 2t-(l/2) if
t is the thickness of the air film at that point and l is the wavelength of
light. At the central point, the lens touches the surface so thickness t=0.
thus the optical path difference is simply l/2, which is the condition for
destructive interference, not constructive interference. so the central spot
has to always be dark.
Q. Why the dark and
bright fringes of Newton’s
experiment are circular?
Ans. The Newton’s
rings are formed due to the phenomenon of thin film interference. here, the
condition for constructive interference(the ring appearing bright) is that the
optical path difference between interfering waves should be an integral
multiple of the wavelength. As the Plano convex
lens is used in Newton’s
rings so the thickness of the film is increasing and then decreasing that’s why
the fringes are circular.
Q.24 Why the
conditions of constructive and destructive interference are reversed in thin
films?
Ans. Reflected light will
experience a 180 degree phase change when it reflects from a medium of higher
index of refraction and no phase change when it reflects from a medium of
smaller index. This phase change is important in the interference which occurs
in thin films, the design of anti-reflection coatings, interference filters,
and thin film mirrors. So in thing film the phase change of 180 degrees occur
that’s why the crests converts into troughs and troughs are converted into
crests. therefore the conditions are reversed.
Q. An oil film over a
wet footpath shows colors? Explain how does it happen?
Ans.This is known as thin-film interference, because it is
the interference of light waves reflecting off the top surface of a film with
the waves reflecting from the bottom surface. To obtain a nice colored pattern,
the thickness of the film has to be on the order of the wavelength of light.
Consider the case of a thin film of oil floating on water. Thin-film
interference can take place if these two light waves interfere constructively:
- the light from the air reflecting off the top surface
- the light traveling from the air, through the oil, reflecting off the bottom surface, traveling back through the oil and out into the air again.
Q. Why Polaroid sun
glasses are better than the ordinary sun glasses?
Ans. There are four things that a good pair of Polaroid sunglasses should do for you:
Sunglasses provide protection from ultraviolet rays in sunlight. Ultraviolet (UV) light damages the cornea and the retina. Good sunglasses can eliminate UV rays completely.
Sunglasses provide protection from intense light. When the eye receives too much light, it naturally closes the iris. Once it has closed the iris as far as it can, the next step is squinting. If there is still too much light, as there can be when sunlight is reflecting off of snow, the result is damage to the retina. Good sunglasses can block light entering the eyes by as much as 97 percent to avoid damage.
Sunglasses provide protection from glare. Certain surfaces, such as water, can reflect a great deal of light, and the bright spots can be distracting or can hide objects. Good sunglasses can completely eliminate this kind of glare using polarization (we'll discuss polarization later).
Sunglasses eliminate specific frequencies of light. Certain frequencies of light can blur vision, and others can enhance contrast. Choosing the right color for your sunglasses lets them work better in specific situations.
Q. Can interference
be without diffraction or vise versa?
Ans. You can have
diffraction without interference. Interference occurs when coherent light waves
coming from two different sources interact. In single-slit diffraction, the two
sides of the slit act as these two sources. If you make the slit much smaller
than the wavelength of whatever you're diffracting, though, it effectively
becomes a single point source, and no appreciable interference occurs and
interference in the thin films is without diffraction.
Chapter #8 :Waves and Sound : Short Q/A / C.R.Q's
Q. If a pendulum is
vibrated at a deep well .What will be the effect on the Time Period?
Ans. As we know that g decreases with the depth.
g’=g
(1-d/ Re 2)
and the
time period of the pendulum is
T=2π√
l/g
So
according to the formula if the value of the g decreases the time period will
increase
Then if a pendulum is vibrated at a deep well its time
period will be grater than on earth.
Q. A wire hangs from
a dark high tower so that the upper end is not visible. How can we determine
the length of the wire?
Ans. First we tied a point mass with this string and then
make it vibrate. The time period of this pendulum can be noticed. Now a/c to
formula
L=gT/4
π2
By knowing
the value of g, the time period of vibration the length of wire can be
calculated.
Q. Is it possible for
two identical waves traveling in the same direction along a string to give rise
to a stationary wave?
Ans. No . A standing wave, also known as a stationary wave, is a wave that
remains in a constant position. This phenomenon can occur because the medium is
moving in the opposite direction to the wave, or it can arise in a stationary
medium as a result of interference between two waves traveling in opposite
directions.
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