Impulse and Momentum Complete Study Notes

By Sidharth Jain|Updated : March 24th, 2022

Complete coverage of the APPSC AE Exam syllabus is a very important aspect for any competitive examination but before that important subjects and their concept must be covered thoroughly. In this article, we are going to discuss the Impulse and Momentum which is very useful for APPSC AE Exams.

Linear Momentum and its Conservation

 

The linear momentum of a particle of mass m moving with velocity ¯v is defined as

 

https://byjusexamprep.com/liveData/f/2017/10/1.PNG-11.PNG

  • Linear momentum, being the product of a scalar and a vector, is a vector.
  • It has dimensions M.L.T–1 and units kg.m.s-1 .
  • The direction of the momentum vector is the same as the velocity vector.

We can express Newton’s second law in terms of linear momentum in this way

https://gs-post-images.grdp.co/2017/10/2-img1509190399740-85.PNG-rs-high-webp.PNG

  • Thus the resultant force on an object (system) equals the time rate of change of linear momentum of the object (system).
  • The definition of linear momentum enables us to put the second law into a more general powerful form.
  • If, in addition, the system is an isolated one then we can formulate a law of conservation of linear momentum.

Impulse and Momentum

  • Impulse is defined as simply change in momentum. In a collision between two particles (and especially a contact collision) the force
    of interaction might vary with time.
  • The force is relatively short-lived, being zero before clock time ti and zero after clock time tf and having a relatively large value at maximum.
  • The elapsed time for the interaction is to a good approximation ∆t = tf – ti.

https://gs-post-images.grdp.co/2017/10/1-img1509195146512-59.PNG-rs-high-webp.PNG

fig above shows A force that varies over a relatively short elapsed time. The area under the force curve is equal to the magnitude of the impulse

  • The sum of forces is a function of time.
  • To find the change in momentum we must integrate over the elapsed time for the interaction:

https://gs-post-images.grdp.co/2018/10/capture-img1540538275998-49.JPG-rs-high-webp.JPG

  • The change in momentum is defined as the impulse and given the symbol ¯J

 

https://gs-post-images.grdp.co/2018/10/capture-img1540537929335-23.JPG-rs-high-webp.JPG

 

  • Impulse has the same dimensions and units as momentum and is also a vector.

  • The impulse has a magnitude equal to the area under the force curve between the two clock times, that is, over the elapsed time of the collision.

  • The direction of the impulse vector is the same as the direction of the change in momentum vector.

By Newton's Second Law, the net force is equal to the mass of object times its acceleration,

 

https://gs-post-images.grdp.co/2018/10/capture-img1540537770537-29.JPG-rs-high-webp.JPG

This can be substituted into the equation for impulse,

 

https://gs-post-images.grdp.co/2018/10/capture-img1540537712601-75.JPG-rs-high-webp.JPG

 

The change in velocity is the difference between the velocities at the starting and ending times,

https://gs-post-images.grdp.co/2018/10/capture-img1540539910757-58.JPG-rs-high-webp.JPGThe formula for impulse becomes,

https://gs-post-images.grdp.co/2018/10/capture-img1540537583530-59.JPG-rs-high-webp.JPG

  • This equation is called the impulse-momentum theorem. In words, it states that the change in momentum of an object in a certain time interval is equal to the impulse of the net force that acts on the object in the time interval.
  • Using this formula, it is possible to relate changes in momentum to the forces that were applied to cause the change.
  • It also shows that the time over which a force is applied has an effect on the change in momentum that results.
  • It is also important to note that the units for momentum and impulse are effectively the same. The unit of momentum is kg m/s, and the unit of impulse is the Newton-second, N.s.

Conservation of Momentum

  • When two objects interact, such as in a collision, they may exert forces on each other.
  • The forces the objects exert on each other can be considered part of a closed or isolated system. In this case, the forces involved are internal forces.
  • If any outside forces affect the system, these are called external forces.
  • According to Newton's Third Law, when there are no external forces, the internal forces that act between two objects have equal magnitudes and opposite directions. If the two objects are labelled A and B, the forces they exert on each other are,

https://gs-post-images.grdp.co/2018/10/capture-img1540538410810-45.JPG-rs-high-webp.JPG

  • During a collision, these forces act for the same amount of time. If the collision begins at time t1 and ends at time t2, then the time duration of the collision is Δt , and the impulse experienced by object A is,

 

https://gs-post-images.grdp.co/2018/10/capture-img1540538891931-49.JPG-rs-high-webp.JPG

  • The impulse experienced by object B is JB,
  • If the values for the forces in these impulse equations are substituted in to the equation for Newton's Third Law, the result is,

https://gs-post-images.grdp.co/2018/10/capture-img1540539092902-64.JPG-rs-high-webp.JPG

By the impulse-momentum theorem, this is equivalent to,

 

https://gs-post-images.grdp.co/2018/10/capture-img1540539477110-47.JPG-rs-high-webp.JPG

  • In this equation, PA,1 means the momentum of object A at time t1 , PA,2  means the momentum of object A at time t2 , PB,1 means the momentum of object B at time t1, and PB,2 means the momentum of object B at time t2.
  • The equation can be rearranged to put all of the terms for time t1 on one side, and terms for time t2 on the other,

https://gs-post-images.grdp.co/2018/10/capture-img1540539554634-73.JPG-rs-high-webp.JPG

  • In this case, in which there were no external forces, the sum of the momenta before the collision is equal to the sum of the momenta after.
  • In general, as long as there are no external forces, the total momentum of the system is constant. This is known as conservation of momentum.
  • For any number of objects  the total momentum can be labeled PA,2,

https://gs-post-images.grdp.co/2018/10/capture-img1540539636925-98.JPG-rs-high-webp.JPG

If there are no external forces, the total momentum P remains constant, even if the moment of the individual objects change.

Instantaneous Impulse

Example: bat and ball contact

5-Impulse_files (2)

  • The relation between impulse and linear momentum can be understood by the following equation.

5-Impulse_files (3)

Where, F = Force, t = time, m = mass, v = initial velocity, u = final velocity

  • Rotation about a fixed point gives the three-dimensional motion of a rigid body attached at a fixed point.

Theory of Collision

  • collision occurs when two or more objects hit each other. When objects collide, each object feels a force for a short amount of time.
  • This force imparts an impulse or changes the momentum of each of the colliding objects.
  • But if the system of particles is isolated, we know that momentum is conserved.
  • Therefore, while the momentum of each individual particle involved in the collision changes, the total momentum of the system remains constant.

The collision between two bodies may be classified in two ways: Head-on collision, and Oblique collision.

  • Head-on Collision
    • Let the two balls of masses m1 and m2 collide directly with each other with velocities v1 and v2 in the direction as shown in figure.
    • After the collision, the velocity becomes and along the same line.

 08-Collisions_files (3)

  • Oblique Collision
    • In case of oblique collision linear momentum of an individual particle do change along the common normal direction.
    • No component of impulse act along the common tangent direction.
    • So, linear momentum or linear velocity remains unchanged along tangential direction. Net momentum of both the particle remains conserved before and after collision in any direction.

General Equation for Velocity after Collision

08-Collisions_files (4)

Where m1 = mass of body 1

  • m2 = mass of body 2
  • v1 = velocity of body 1
  • v2 = velocity of body 2= velocity of body 1 after collision= velocity of body 2 after collision

Where e = coefficient restitution

  • In case of head-on elastic collision e = 1
  • In case of head-on inelastic collision 0 < e < 1
  • In case of head-on perfectly inelastic collision e = 0

The procedure for analyzing a collision depends on whether the process is elastic or inelastic. Kinetic energy is conserved in elastic collisions, whereas kinetic energy is converted into other forms of energy during an inelastic collision. In both types of collisions, momentum is conserved. 

  • Elastic Collisions
    • Some kinetic energy is converted into sound energy when pool balls collide otherwise, the collision would be silent and a very small amount of kinetic energy is lost to friction.
    • However, the dissipated energy is such a small fraction of the ball’s kinetic energy that we can treat the collision as elastic.

Equations for Kinetic Energy and Linear Momentum

  • Let’s examine an elastic collision between two particles of mass http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total304.gif and http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total305.gif, respectively. Assume that the collision is head-on, so we are dealing with only one dimension—you are unlikely to find two-dimensional collisions of any complexity on SAT II Physics. The velocities of the particles before the elastic collision are http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total306.gif and http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total307.gif, respectively. The velocities of the particles after the elastic collision are http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total308.gif and http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total309.gif. Applying the law of conservation of kinetic energy, we find:

http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0001/12m1v1.gif

 

Applying the law of conservation of linear momentum:

http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0009/m1v1+m2v2.gif

 

  • These two equations put together will help you solve any problem involving elastic collisions. Usually, you will be given quantities for http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total310.gifhttp://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total311.gifhttp://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total312.gif and http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total313.gif, and can then manipulate the two equations to solve for http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total314.gif and http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total315.gif
  • A head-on with the cue ball in pool, Both of these balls have the same mass, and the velocity of the cue ball is initially http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0002/phy.total.SMALL316.gif. What are the velocities of the two balls after the collision? Assume the collision is perfectly elastic

https://gs-post-images.grdp.co/2017/10/1-img1509344788912-18.PNG-rs-high-webp.PNG
Substituting http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total317.gif and http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total318.gif into the equation for conservation of kinetic energy we find:

http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0001/12mv21.gif

Applying the same substitutions to the equation for conservation of momentum, we find:

http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0009/mv1=mv1.gif

If we square this second equation, we get:

http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0015/v21=v1r2.gif

By subtracting the equation for kinetic energy from this equation, we get:

http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0002/2v1v2-xdf.gif

 

  • The only way to account for this result is to conclude that http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total319.gif and consequently.http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total320.gif

Inelastic Collisions

  • Most collisions are inelastic because kinetic energy is transferred to other forms of energy—such as thermal energy, potential energy, and sound—during the collision process.
  • The kinetic energy is not conserved, in inelastic collision. Momentum is conserved in all inelastic collisions.
  • The one exception to this rule is in the case of completely inelastic collisions.
  • Completely Inelastic Collisions
    • A completely inelastic collision also called a “perfectly” or “totally” inelastic collision, is one in which the colliding objects stick together upon impact.
    • As a result, the velocity of the two colliding objects is the same after they collide.
    • Because http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total321.gif, question may be asked for finding

for eg. below two gumballs, of mass m and mass 2respectively, collide head-on. Before impact, the gumball of mass m is moving with a velocity http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0002/phy.total.SMALL322.gif, and the gumball of mass 2m is stationary.

https://gs-post-images.grdp.co/2017/10/1-img1509345616321-69.PNG-rs-high-webp.PNG

  • First, note that the gumball wad has a mass of m + 2m = 3m. The law of conservation of momentum tells us that http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total324.gif, and so http://img.sparknotes.com/content/testprep/bookimgs/sat2/physics/0017/phy.total325.gif. Therefore, the final gumball wad moves in the same direction as the first gumball, but with one-third of its velocity.

 

 

08-Collisions_files (8)

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