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



1
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? 

2
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? 

3
To the nearest tenth of a Joule, what is its maximum kinetic energy of the pendulum in problem 2? 

4
To the nearest tenth of a Joule, what is the total energy of the pendulum in problem 2? 

5
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? 

6
If the slope in the above problem is 23 degrees, what is the coefficent of friction (to 2 decimal places)? 

7
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? 

8
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? 

9
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? 

10
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? 


Answers

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.

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:
  1. the light from the air reflecting off the top surface
  2. 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.