QUESTIONS FOR SHORT ANSWERS
1. Galileo was punished by the Church for teaching that the sun is stationary and the earth moves around it. His opponents held the view that the earth is stationary and the sun moves around it. If the absolute motion has no meaning, are the two viewpoints not equally correct or equally wrong?
Answer: As far as absolute motion is concerned the two viewpoints are equally correct or equally wrong. Galileo is true only if the point of reference is chosen away from the solar system like another star.
2. When a particle moves with constant velocity, its average velocity, its instantaneous velocity, and its speed are all equal. Comment on this statement.
Answer: A particle moving with constant velocity move an equal distance in equal time whatever the magnitude of time taken to compare. So the ratio of distance to time is always constant. So the average velocity (which is the ratio of total displacement to total time), instantaneous velocity (Which is the ratio of infinitesimal distance covered to infinitesimal time at the instant) and speed (which is the ratio of distance covered to the time taken) are all equal.
3. A car travels at a speed of 60 km/hr due north and other at a speed of 60 km/hr due east. Are the velocities equal? If no, which one is greater? If you find any of the questions irrelevant, explain.
Answer: The velocities are not equal because velocity is a vector quantity having both magnitude and direction. In this case, both are equal in magnitude but not in direction. The question "which one is greater?" is irrelevant because two vectors can only be compared if they have the same direction.
4. A ball is thrown vertically upward with a speed of 20 m/s. Draw a graph showing the velocity of the ball as a function of time as it goes up and then comes back.
Answer: When the ball goes up its velocity at any instant is given by
v=u-gt, here u=20 m/s, let g= 10 m/s². this gives v=20-10t; on the time-velocity graph, it is an equation of a straight line with an intercept of 20 m/s on the Y-axis (along which velocity is plotted) and gradient of '-10'. Let us find time t for the v to become zero,
0=20-10t
→ t=2 seconds, so this straight line will touch the X-axis (along which time is plotted) at t=2 seconds.
When the ball goes down its velocity at any instant reverses its direction ie it becomes negative and in the next 2 s it comes back to the initial point with 20 m/s. So it can be plotted as below:-
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| The graph for question no- 4 |
5. The velocity of a particle is towards the west at an instant. Its acceleration is not towards the west, not towards the east, not towards the north and not towards the south. Give an example of this type of motion.
Answer: Consider a case when a particle is thrown towards the west at an angle with horizontal. At the highest point of its trajectory, its instantaneous velocity is towards the west but the acceleration is not towards the west, not towards the east, not towards the north and not towards the south. In fact, its acceleration is vertically downwards due to gravity.
6. At which point on its path a projectile has the smallest speed?
Answer: At the highest point on its path a projectile has the smallest speed.
7. Two particles A and B start from rest and move for equal time on a straight line. The particle A has an acceleration a for the first half of the total time and 2a for the second half. The particle B has an acceleration 2a for the first half and a for the second half. Which particle has covered a larger distance?
Answer: Let '2t' be the total time for which the particles move, vA and vB the velocities of particle A and B after the first half of the total time. From the equation v=u+at,
For particle A, u=0, vA = at
using equation s=ut+1/2 at²
Total distance traveled by particle A
=(ut+½at²)+(vAt+½2at²)
=(0+½at²)+(at.t+at²) =½at²+2at²
=2.5 at²
For particle B, vB=0+2at =2at,
So total distance traveled by particle B
=(ut+½2at²)+(vBt+½at²)
=(0+at²)+(2at.t+½at²)
=at²+2at²+½at² =3at²+½at²
=3.5 at²
Comparing both it is clear that particle B covered a larger distance.
8. If a particle is accelerating, it is either speeding up or speeding down. Do you agree with this statement?
Answer: This statement cannot be agreed upon. Consider the case of a uniform circular motion where acceleration is always perpendicular to the path of motion. Here the particle neither speeds up nor speeds down but moves with constant speed.
9. A food packet is dropped from a plane going at an altitude of 100 m. What is the path of the packet as seen from the plane? What is the path as seen from the ground? If someone asks "What is the actual path", what will you answer?
Answer: The path of the packet as seen from the plane is a straight line. The path of the packet as seen from the ground is parabolic.
The actual path has no meaning, we can describe the path only with respect to some reference point.
10. Give an example where (a) the velocity of a particle is zero but its acceleration is not zero, (b) the velocity is opposite in direction to the acceleration, (c) the velocity is perpendicular to the acceleration.
Answer: The followings are the examples:-
(a) A particle thrown vertically upwards has its velocity zero at the highest point but its acceleration is not zero. The acceleration is 'g' due to gravity vertically downwards.
(b) When a particle is thrown vertically upwards its velocity at any instant during upwards movement is opposite in direction to the acceleration. Velocity is vertically upwards while acceleration due to gravity is vertically downwards.
(c) In a uniform circular motion, the velocity is perpendicular to the acceleration.
11. Figure (3-Q1) shows the x- coordinate of a particle as a function of time. Find the signs of vx and ax at t=t1, t= t2 and t=t3.
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| H C Verma's Concepts of Physics-Part 1, Chapter 3-REST AND MOTION: KINEMATICS"-'Questions for short answer' |
Answer: If x' and x" be the x-coordinate of the particle at initial time t' and t" respectively then vx =(x"-x')/(t"-t') = tan Θ. For t"-t' infinitesimally small it is the vx at that instant. So the slope of the tangent at any point in the above graph gives vx. At t=t1, tan Θ is positive, so sign of vx is positive. At t= t2 the slope of the curve is horizontal, so tan Θ=0 → vx =0. At t=t3 the slope of the curve is negative, so sign of vx is negative.
Sign of aₓ
Acceleration is the rate of change of velocity. For the sign of the acceleration at any instant, we see the slope of the curve (which is velocity) before and after the instant. If the slope after the instant is more than before the instant, it means the velocity is increasing i.e. the acceleration is positive, otherwise, it is just opposite.
At t = t₁, the slope is increasing. Hence the sign of aₓ is positive. At t = t₂, the slope is less after the instant than it was before, so the velocity is decreasing, hence the sign of aₓ is negative here. At t = t₃, the slope before the instant is less while after the instant it is more, so the velocity is increasing, hence the acceleration (aₓ) is positive.
12. A player hits a baseball at some angle. The ball goes high up in the space. The player runs and catches the ball before it hits the ground. Which of the two (the player or the ball) has greater displacement?
Answer: Both have equal displacements because both (the player and the ball) start from the same point and stop to the same point. Even though their paths are different the displacement is the vector joining initial to the final position.
13. The increase in the speed of a car is proportional to the additional petrol put into the engine. Is it possible to accelerate a car without putting more petrol or less petrol into the engine?
Answer: It is possible to accelerate a car without putting more petrol or less petrol into the engine if it is driven down a hilly road. The slope of the road should be more than enough to overcome the force of friction.
14. Rain is falling vertically. A man running on the road keeps his umbrella tilted but a man standing on the street keeps his umbrella vertical to protect himself from the rain. But both of them keep their umbrella vertical to avoid the vertical sun-rays. Explain.
Answer: The tilt of the umbrella will be in the direction of the relative velocity of the rain or sun-rays to the man. The relative velocity of the rain or sun-rays to the man will be the velocity of rain or sun-rays minus the velocity of man. For a standing man, this relative velocity will be vertical because his own velocity is zero so he keeps his umbrella vertical for both the rain or sun-rays.
For the running man in the case of rain this relative velocity is at an angle to the vertical, so to protect himself he keeps his umbrella tilted in the direction of the relative velocity. In the case of sun-rays the velocity of light is so much greater that man's velocity is negligible. So the relative velocity of sun-rays is practically vertical so the running man keeps his umbrella vertical.
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CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 18 - Geometrical Optics
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
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Click here for → Exercise(43-54)
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CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 6 - Friction
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
i am very thankfull to you for this solutions
ReplyDeleteWelcome! All the best.
DeleteQuestion fifth is wrong as vertically downwards is south
ReplyDeleteDear Student, I saw your comment today.
DeleteWhen you hang a map on the wall then vertically downward is South but in reality vertically downward is not the South.
When you stand on the ground facing East - your back is towards West, left hand towards North and the right hand towards the South. The vertically downward is towards your feet (as referred in the said answer) and vertically upward is just above your head.
Hope this clears the doubt.
Thanks a lot sir ....i think u have missed the signs for acceleration in Q 11....according to me the signs would be +ve at t1 , 0 at t2 and -ve at t3 ...please coorect me if wrong.
ReplyDeleteI have added the signs for accelerations. But please see the answer with explanation for the signs of accelerations.
DeleteSir I was having doubt in Q 11 . Sir first the velocity was positive then it became zero and then it became negative .doesn't this mean that the acceleration was negative all the time ?
ReplyDeleteDear student, we are dealing with the instant accelerations. Your doubt may be due the projectile motion and its path. The path is convex everywhere from above. And at each point the slope continuously decreases, hence the acceleration is always negative. It is not the case here.
DeleteThank you sir
ReplyDeleteSir,in 11th ques, should the acceleration not be negative as the slope is decreasing?
ReplyDeleteand thanks a lot for the solutions sir!
Dear student, sorry for the late response.
DeleteAt time t3, the slope is negative, when the numeric value of a negative entity decreases, the entity increases. So the slope here is actually increasing.
Hope it is clear. Feel free to ask if any doubt.
Thank you.