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Center of Mass, Momemtum, Collision
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Center of Mass Numericals

PROBLEMS (EASY TYPE)

1) A wooden plank of mass m0 is resting on a smooth horizontal floor. A man of mass am0 starts moving from one end of plank to other. The length of plank is l0. Find the displacement of plank over the floor when man reaches other end of plank?

Solution:


The center of mass remains stationary as no external force. Hence,



2) A particle of mass 5kg is initially at rest. A force starts acting on it one direction whose magnitude changes with time. The force time graph is as shown in figure. Find the velocity of particle after 10 sec?


Solution:

Impulse = DMomentum

             = Area under F-t graph.


3) Three identical balls, ball A, ball B, ball C are placed on a smooth floor on a straight line at the separation of 20m between balls as shown in figure. Initially balls are stationary. Ball A is given a velocity of 5m/s towards Ball B. Collision between A+B has coefficient of restitution as ½. But collision between B+C has coefficient of restitution as 1. What is the time interval between 2 consecutive collision between ball A and ball B?


Solution:

(1) Collision of A and B:

If velocities of ball A and ball B are V1 and V2 after collision then,

Then

 

 


And


(2) Collision between B and C:

Velocities get interchanged and so B is at rest after collision.

So time interval =

                          = 16 sec

4) A light spring of spring constant K0 is kept compressed with compression =

between two masses of mass ‘m’ and ‘am’. When released the blocks acquire a velocity in opposite directions. The spring loses contact with both blocks when it comes in natural length! Find the final velocities of the two blocks?

Solution:


Since no external force is acting, hence

But initial momentum = 0. Hence if final velocities of m and am are V1 and V2 then,

mV1 - amV2 = 0

V1 = aV2 ------ (1)

Now energy remains conserved so,

Ei = Ef.


5) A bullet of mass m0 strikes a block of mass ‘hm0’ with a speed of V0 and gets embedded into the mass ‘hm0’. The block is attached to a spring of stiffness ‘K’. Find the loss of K.E. of system after impact?

Solution:

                                   

As the spring force is a non impulsive force, so we can conserve momentum as:

         

 

 

6) A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other with same K.E =E0 Find energy of explosion?

Solution:

Let the three fragments move along x, y, z direction with velocities .

Now




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  Electricity and Effect of Current      Centre of Mass      Work Power & Energy      Magnetic Effects of Current      Introduction to Laws of Motion      AIEEE 2009 Solutions and Analysis      Chemical Bonding      Quadratic Equations      Progression And Series      dasda      Chemical Bonding      General Principles of Extraction of Metals      Ether and epoxides      Atomic Structure      Parabola      IIT-JEE Chemistry by M.K.Tiwri      Properties of Matter      Heat & Thermodynamics      Aldehydes and ketones      Disha - Path to Success Series     
 



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