Showing posts with label One Dimensional Motion. Show all posts
Showing posts with label One Dimensional Motion. Show all posts

Equations of Motion for Vertically Thrown up body

Equations of motion for a vertically thrown up body

Thus the velocity of the body keeps on decreasing and at a particular height, its velocity becomes zero and that height is said to be the maximum height that the body can go and the time taken to reach that maximum height is called time of ascent. 

Acceleration in this case is acceleration due to gravity which is constant and it shall be treated as negative as the velocity of the body keeps on decreasing with respect to time. Taking this into consideration, we can rewrite the four equations of motion for a freely falling body. It is proved that the velocity with you project the body vertically up is the velocity with which it comes back to the same point of projection but in the opposite direction. It is also proved that when air resistance is ignored, time of ascent is equal to the time of descent.




 Time of Ascent and Decent with Air resistance

As it is mentioned earlier time of ascent of a vertically thrown up body is equal to the time of descent of a freely falling body. The above mentioned statement is true only when air resistance is ignored. But in real life air resistance is there and if that is taken into consideration, time of ascent is different from time of descent. Air resistance offers a force against the motion and it always acts opposite the motion. Thus when the body is thrown up, both gravitational force and resistance force acts against the motion. When the body is coming down, gravitational force acts in down ward direction but air resistance acts opposite to the motion and that is in upward direction. Thus we can find the effective acceleration in both the cases and find the the relations between time of ascent and descent as shown in the video below.




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Motion in a Straight line Introduction and Average Velocity Video Lesson

Motion in a straight line an introduction

Studying the motion of the body without bothering about the forces acting on it is done in kinematics. We treat body as a combination of identical point sized objects and they have negligible dimensions. All laws of mechanics were in principle discussed with the point sized particles and as the body is the combination of similar particles, under ideal conditions the laws are applicable to bodies also. Here we are dealing with bodies moving with a velocity much lesser than the velocity of the light. In this particular case, body is moving only along one dimension either along X,Y or Z axis. This is called one dimensional motion and it is changing its position with respect to time and surroundings.

To measure the change of the position, we have terms like distance, speed. Distance is the actual path traveled by a body and the speed is the rate of change of distance with respect to time. Both distance and speed are treated as scalars and they can be understood by stating their magnitude alone and they don’t need direction.

Displacement is the shortest distance between initial and final positions in specified direction and it is treated as vector quantity. They can be understood completely only when both magnitude and directions are given to us. Velocity is defined as the rate of change of displacement and it is also a vector quantity.




 Average velocity

If a particle is not changing its velocity with respect to time, then it is said to be in uniform velocity. In this case at any given interval of time, the particle will have same constant velocity and it is same every where. But it is not same every where. If a body is changing its velocity with respect to time, then it is having acceleration and we would like to measure the average velocity in the given case. Average velocity is defined as the ratio of total displacement covered by a body in the total time. Taking this concept into consideration, we can find average velocity when time is shared and displacement is shared as shown in the video below.




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Collisions Problems and Solutions Two

One dimensional elastic collision means bodies before and after collision travel in the same direction.  If both momentum and kinetic energy are conserved, the collision is said to be elastic collision. There is no wastage of energy in this case and all kinetic energy is conserved. In the case of one dimensional elastic collision, the velocity of approach of two bodies before collision is equal to the relative velocity of separation after the collision.This ratio is called coefficient of restitution and its value for elastic collision its value is one. In the case of perfect inelastic collision, the two bodies move together as one body and they have same velocity.

 Collisions is a phenomenon where energy and momentum between two bodies are in interaction. There need not be physical contact for the collision. Change of path and transfer of linear momentum and kinetic energy is sufficient to say that the collision is happened. In the case of elastic collision, both kinetic energy and linear momentum are conserved. In the case of inelastic collision, only linear momentum is conserved but not kinetic energy. Some part of kinetic energy is converted into other formats of energy in this case. In the case of perfect inelastic collision, both the bodies after the collision move together and will have a common velocity.

Problem

A particle of mass one kilogram is thrown vertically upward with a speed of 100 meter per second.After five seconds it explodes into two parts of different masses. If velocity of one particle is known, we need to measure the velocity of the other part. Problem is as shown in the diagram below.


Solution

The body is moving against the acceleration due to gravity and we can find its velocity after five seconds using first equation of motion. We can also apply law of conservation of linear momentum and find the velocity of the second particle by applying proper sign convention. The solution is as shown in the diagram below.


Problem

It is given in the problem that a projectile of mass m is projected at an angle to the horizontal. At the maximum height of the projectile it breaks into two fragments of equal masses. One of the fragment retraces its path and we need to measure the velocity of the other piece.


Solution

At the maximum height of the projectile it has only horizontal component of the velocity and its vertical velocity component at that instant is zero. We can measure the initial momentum of the particle at the maximum height. The retraced particle has the same velocity but in the opposite direction. We can apply law of conservation of linear momentum and solve the problem as shown in the diagram below.


Problem

A particle of mass is moving with a velocity and it collides head on with another stationary particle and the collision is elastic. If the velocity of the second particle is known, we need to find the ratio of the mass of the two particles. Problem is as shown in the diagram below.


Solution

We can apply first law of conservation of linear momentum and find the relation between first and second velocities. We can also apply the concept of velocity of approach is the velocity of separation and get on more result. The solution is as shown in the diagram below.


Problem

It is given in the problem that a bullet of mass m moving with a velocity strikes a suspended wooden block of mass M. If the block rises to a height h, we need to measure the initial velocity of the bullet. Problem is as shown in the diagram below.


Solution

We can apply law of conservation of linear momentum to the given scenario and find the common velocity of the system when bullet is embedded into the block. This kind of collision is called perfect inelastic collision. We can equate the kinetic energy of the system into potential energy of the system and it can be further simplified as shown in the diagram below.


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Collisions Problems with Solutions One

Collision is a physical action between two bodies where they exchange momentum and kinetic energy. There is no rule that the bodies need to come into contact for this to happen. Even with out physical contact, if the bodies of the system exchange their momentum and kinetic energy. For example  an alfa particle striking towards a heavy nucleus, they deviate from their path and it also shall be treated as collision . If both linear momentum and kinetic energy are conserved, the collision is said to be elastic collision. If only linear momentum is only conserved and some of the kinetic energy is converted into light, heat or sound, that kind of collision is called inelastic collision.

Problem

A stationary shell explodes into two fragments with masses having ratio 1 : 2. If the kinetic energy of the heavier piece is 100 joule, we need to measure the kinetic energy released in the explosion.


Solution

As the body is initially in the state of rest, its initial linear momentum is zero.As momentum is conserved, by applying it we can get the ratio of velocities of the bodies as shown in the diagram below. With knowledge about both velocity and mass, we can find individual and total kinetic energy of the body as shown in the diagram below.


Solution

A uranium nucleus at rest emits an alpha particle with a velocity known. We need to find the recoil velocity of the remaining particle. Problem is as shown in the diagram below.


Solution

As there is no external force on the atom, linear momentum is conserved. As we know the mass of two parts, we can apply conservation of linear momentum and solve the problem as shown in the diagram below.


Problem

A particle of mass 3m is moving with a velocity and it has elastic collision with another particle of mass 2m which is at rest. We need to measure the final velocities of the bodies after collision. Problem is as shown in the diagram below.


Solution

We shall know the formulas to find the final velocities of the two bodies after one dimensional elastic collision and they are as mentioned below. Problem is further solved as shown.


Problem

A moving particle of mass m makes a straight collision with another particle of mass 4m which is at rest. We need to know the fraction of the kinetic energy retained by the incident particle and the problem is as shown in the diagram below.


Solution

As the collision is one dimensional elastic collision, coefficient of restitution is equal to one. So the ratio of velocity of separation after the collision is equal to the velocity of approach before the collision. Also by applying conservation of momentum, we can get one more equation and by simplifying them , we can solve the problem as shown in the diagram below.


Problem

A gun of mass M fires a bullet of mass m with a kinetic energy E. We need to measure the velocity of recoil of the gun.


Solution

By applying law of conservation of linear momentum, we can find the velocity of the second body as shown below. By substituting in the formula of kinetic energy, we can solve the problem as shown here.


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Motion in One and two Dimensions Problems with Solutions Fourteen

We are solving problems on one and two dimensional motion in this series of chapters. Here we are dealing with motion of a body moving along only one dimension and also we are studying the motion of a body in a plane where it has motion both along X and Y axis simultaneously.

In the given problem a body is dropped from a certain height and it strikes the inclined plane at a height h above the ground. As a result of impact, velocity of the body becomes horizontal. We need to measure  that it will take maximum time to reach the ground basing on what condition. The problem is  as shown in the diagram below.


Solution

It is given that the body is dropped from a certain height and hence its initial velocity is zero. It has fallen only some distance before it actually strikes the ground. That can be found as the difference between the total height from where it  is dropped and the location where it hit the ground.

Thus we can measure the time of fall in both the cases using the equations of motion. We need to measure the total maximum journey time. For the parameter to be maximum, its differentiation function with height has to become zero as a mathematical rule. 

By simplifying it, we can find that relation as shown in the diagram below.


Problem

It is given in the problem that a projectile is projected with a velocity and a known angle of projection. It is given that the trajectory grazes the vertices of the triangle as shown in the diagram. We need to know the relation between the angles with the horizontal.



Solution

The diagram for the problem data is as shown in the figure. We can express the angles with the horizontal in terms of horizontal and vertical displacement as shown in the diagram below.

We also know that the vertical displacement can be expressed in terms of horizontal displacement and angle of projection in terms of range as shown further.

By comparing this two equations, we can find the relation as shown below.


Problem

Two cannons are installed at the top of a cliff of 10 meter and they fire a shot with a known speed at some interval of time. The first cannon is fired with an angle and the other one is done horizontally. We need to know where do the two cannons collide with each other. The problem is as shown in the diagram below.


Solution

Both of them were fired from a certain height. But one horizontally and the other with an angle. By substituting the given data, we can simplify the equations as shown in the diagram below.



Problem

It is given in the problem that a particle is projected from the ground with a initial speed at angle of projection with the horizontal. We need to measure the average velocity of the projection between its point of projection and the highest point of the trajectory. The problem is as shown in the diagram below.


Solution

We know that the average velocity is the ratio of total displacement of a body in the total time of the journey. We can find the total displacement of the journey as a vector sum of horizontal and vertial displacements and the time of journy is half of time of flight.By substituting that data, we can  solve the problem as shown in the diagram below.



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Motion in One Dimension Problems with Solutions Thirteen

We are solving series of problems in one dimensional motion. It is given in the problem that a point moves in a straightline under a retardation as given in the problem where k is a constant.Acceleration is defined as rate of change of velocity. If a body has acceleration means, its velocity is increasing with time. If it is deceasing with  time, it is called retardation.

It is given that the particle has initial velocity. We need to know the distance covered by it in the given time. The problem is as shown in the diagram below.


Solution

We know that retardation is a negative acceleration. Here in this case the velocity of the body is decreasing with time. To indicate it, a negative sign is taken into consideration. We can rearrange the velocity  in terms of time as shown in the diagram below. As we know the part of velocity, to get the total velocity by integrating it in the given limits. By using rules of integration, we can the value of final velocity after applying the limits of lower and upper.

We need displacement and hence write the velocity as the rate of change of displacement. We can write the equation for small displacement and by integrating it, we can get the displacement of the particle as shown in the diagram below.


Problem

In the arrangement shown in the figure and instantaneous velocities of two masses, we need to know the angle between the vertical as shown in the diagram below. 


Solution

Let the vertical distance between pulley and the surface is X. Let the horizontal distance is 2a. It is given that the length of the string is constant and hence it can be found in the given terms as shown in the diagram below. As the equation is constant, its differentiation with time is zero.

By applying that condition, we can the relation between two velocities as shown. Basing on that we can find the angle and the detailed solutions is given below.


Problem

It is given in the problem that a ball is thrown vertically upward with a known velocity. While going upward and again while coming back after reaching its maximum height, the ball came back to the same point and the time interval between them is given to us in the problem. We need to measure the initial velocity of the ball. The problem is as shown in the diagram below.


Solution

Let the height that is crossed twice is and the intervals are given in the problem as shown in the diagram. Thus we can equate the height equation both in terms of the first time and the second time. By solving these two  equations, we can solve the problem as shown in the diagram below.


Problem

It is given in the problem that a body is projected vertically and different points are given in the journey. If the body is released from position A and we need to know the time of descents ratio need to be found.


Solution

We know that the time intervals for each part is same and hence to reach each point, the time taken in multiples of that interval as shown in the diagram below. We need to use the second equation of motion and solve the problem as shown in the diagram below.


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Motion in One Dimension Problems with Solutions Twelve

We are interested in solving the problems based on displacement,velocity and acceleration. They are all belongs to one dimensional motion. We know that the velocity is the rate of change of displacement and acceleration is the rage of change of velocity.

In the first problem it is given that position of the particle is given in the XY plane and it is given in terms of time and trigonometric function. We need to know the path of the particle and the problem is as shown below.


Solution

Position vector is given to us in terms of both X and Y components. By multiplying with I unit vector we identify it as the X component. Similarly by multiplying with J, we can identify the component as Y component. J is a unit vector along Y axis. Unit vector is a vector with specific direction and has magnitude of only one unit.

Using trigonometric rule, we can find the relation between X and Y as shown in the diagram below. The path is a circle as the equation supports that mathematically.


Problem

A particle moves according to the equation given in the problem with some constants. We need to find the velocity of the particle in terms of time and we can assume that the body starts from rest. The problem is as shown in the diagram below.


Problem

Rate of change of velocity is given in the problem and we can rearrange the terms to get the component of velocity. But we need the total velocity and to get that we shall integrate the component of velocity with certain given limits. By simplifying the equation further, we can solve the problem as shown in the diagram below.


Problem

A particle moves in a straight line and the position of the particle is given to us in terms of time as shown in the diagram below. We need to know how the velocity and acceleration of the particle changes with respect to time and we need to check with the options given in the diagram below.


Solution

What is given in the problem is in terms of displacement and to get the velocity we need to differentiate the equation. It can be found further that after two seconds velocity of the particle and its displacement becomes zero.

By differentiating the velocity further we can get acceleration as shown below. It can be concluded basing on the equations that if time is less than three seconds, acceleration could be negative.


Problem

It is given in the problem that after the engine is switched off, the boat goes for retardation and it is given as shown in the diagram below. We need to know the velocity of the particle after a specified time.


Solution

By rearranging the terms, we can get a part of velocity. To get the total velocity in the given limits, we need to integrate the equation. Thus we get the velocity of the particle at a given time as shown in the diagram below.



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