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Shilpa posted a question via the contact form available here:
“Can you suggest me some ways and books to make my physics better?”
She was considering the question as a silly one. But this is a serious question. Many of us do not ask many questions which come into our mind out of the thought that it may be silly or foolish.
“Only those questions are foolish which are not asked”
As the motto of our sister project “AskPhysics” goes.
How to make Physics better?
Here actually the question is not to make Physics better, but it is to make the performance and understanding in Physics better.
Making the fundamentals strong.
The basic principles of Physics must be grasped right from the grassroot level and must be assimilated through practice of related problems. The wide range of phenomena and applications coming under Physics are all based on a handful of laws and principles.
Physics is basically a problem solving discipline and must be taken in that spirit. After learning a concept try to solve as much problems related to it as possible. This well help you to apply what you have learnt and will encourage you to think in multiple dimensions thereby developing your brain functioning. This will be of great help to solve problems of real help too. A person with strong problem solving ability in Physics will be successful in life too. (In handling and solving problems)
Visualize what you learn:
Do not try to learn things as mere “text”. Grasping Physics requires lot of imagination. When you learn something, try to visualize it. Draw diagrams to understand things better.
In learning (particularly Physics), “Questioning every answer is more important than answering every question“. Think deeply on whatever you learn and try to question the new knowledge. Dare asking doubts to your teachers. (If you cannot, post them at www.askphysics.com or http://physicsforum.plustwophysics.com)
Collect and solve previous question papers.
This is important to get a better score in exams.
For developing a strong foundation in Physics.
- Conceptual Physics: Paul G Hewitt
- Fundamentals of Physics : Hallidey Resnick Krane
For Problem Solving
- Concepts of Physics: HC Verma
- Problems in Physics : IE Irodov
Here are some links to further exploring “How to Study Physics?”
Irodov Problems in Physics is a set of high standard problems in Physics
FInd the solutions to these problems here
The body mpushes the body Mbackwards. As the bodyM is pushed backwards,m is forced to slide down the incline on M. Let the tension in the string be T and let the normal reaction between the surfaces be N. Further let the acceleration of body M be and let the acceleration of body m with respect to an observer on body Mbe along the direction down the inclined plane.
Relation between and : The relation between the two acceleration can be seen in the figure below. As seen in the figure,AB and CD are two section of the string before and after the pulley. After the mass M and hence the pulley moves back by x units the length CD shortens to C’D = CD-x. Since, the total length of the stringAB + CD is to remain constant AB must extend to A’B’ = AB + x. In other words if the mass M moves x units towards the wall, the massm slides the same x units on the inclined plane. Thus, we have,
Forces on mass m : We will resolve the forces acting on m in the parallel and perpendicular direction to the incline. This mass experiences two kinds of i) accelerations, as it rides along with mass M and ii) its acceleration as it slides on the incline relative to M. The net acceleration is the summation of these two accelerations. From (1) however, the magnitude of both these accelerations is the same.
In the direction perpendicular to the incline, there are twp forces acting on the body, i) the component of gravity and ii) the normal reaction N from the surface of mass M. The component of the net acceleration of mass m along the perpendicular direction is given by, as shown in the figure. Thus we have,
In the direction parallel to the incline, there are two forces acting on the mass m, i) the tension in the string T and ii) the component of force of gravity pulling it down the incline. The component of net acceleration along this direction is given by as shown in the figure. Thus, we have,
Forces on mass M : For this problem we need consider only the forces acting in the horizontal direction – this is shown in the figure.
There are three forces acting on the mass M that effect its motion in the horizontal direction, i) the normal reaction from mass mand ii) the tensions of magnitude T in the parts of the string after and before the pulley directed along the direction of the string. The component of tension in the part string connecting the mass m and the pulley is given by . The mass M accelerates at a rate w towards the wall. Thus we have,
Now we have all the information needed to solve for w.
From (2) and (4),
From(3) and (5) we have,
In the system, as body 1 moves backwards, body 2 will slide down along the inclined plane. Let the acceleration of body 1 be and let the acceleration of body 2 with respect to an observer on body 1 be along the direction down the inclined plane. Further let the normal reaction between the two bodies be N.
Forces acting on body 2 : We shall resolve all forces in directions parallel and perpendicular to the incline of body 1. There are two forces acting on this body in the perpendicular direction.
i) the component of force of gravity and ii) the normal reaction N between the surfaces. The only acceleration experienced by the body in this direction is the component of its acceleration as it rides on body 1 – as shown in the figure. Thus, we have,
In the direction parallel to the inclined plane, there is only one force acting on the body – the component of force of gravity that’s pulling it down the inclined plane, . The net acceleration of the body along this direction is sum of two accelerations it is being subject to, i) its acceleration as it rides on body 1 – and ii) its acceleration relative to an observer on body 1 along the inclined plane . Thus, we have,
Forces acting on body 1: We shall consider only the horizontal direction for this part. There is only one force acting on the body – the component of normal reaction that is responsible to accelerate the body at a rate . Thus, we have,
From (1) and (3)