- Introduction
- |
- Center of Mass
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- Position of center of mass
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- Position vector of centre of mass in terms of co-ordinate components
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- Motion of centre of mass
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- Acceleration of centre of mass
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- Kinetic energy of the system of particles
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- Two particle system and reduced mass
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- Collisions
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- Head on elastic collision of two particles
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- Head on inelastic collision of two particles
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- Deflection of an moving particle in two dimension
- Solved examples

- Included with Linear momentum
- Assignment 1
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- Assignment 2
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- Assignment 3
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- Assignment 4

- Included with Linear momentum
- How to solve center of mass problems

**A) In laboraratory frame of refrence**

- Consider two particles whose masses are m
_{1}and m_{2}.Let**u**and_{1}**u**be the respective velocities before collision_{2}

- Let both the particles stick together after collision and moves with the same velocity
**v**.Then from law of conservation of linear momentum

m_{1}**u**+m_{1}_{2}**u**=(m_{2}_{1}+m_{2})**v**

or

- If we consider second particle to be stationary or at rest then
**u**=0_{2}

then

m_{1}**u**=(m_{1}_{1}+m_{2})

or

Hence |

**v**| < |**u**|_{1}

- Kinetic energy before collision is

KE_{1}=(1/2)m_{1}u_{1}^{2}After collison KE of the system is

KE_{2}=(1/2)(m_{1}+m_{2})v^{2}

Hence from equation we come to know that K

_{2}< K_{1}hence energy loss would be there after thye collision of the particles

**B) In center of mass frame of refrence:-**

- Velocity of center of mass is

when second particle is at rest.

It is clear that Center of masss frame of refrence moves with velocity v_{cm}w.r.t laboratory frame of refrence.

Hence velocity of particle having mass m_{1}in C-frame of refrence is

And velocity of particle having mass m

_{2}in C-frame of refrence would be

- let
**v**^{'}be the joint velocity of the particles which stick to each other after the collision in center of mass frame of refrence .So by the law of conservation of linear momentum(m

_{1}+m_{2})**v**^{'}=m_{1}**u**+m_{1}^{'}_{2}**u**_{2}^{'}Substituting the values of

**u**and_{1}^{'}**u**we find_{2}^{'}

(m_{1}+m_{2})**v**^{'}=0

but (m_{1}+m_{2}) can be zero so

**v**^{'}=0Therefore after collision total momentum of the particles comes out to be zero.Because both the particles stick to each other after collision so the velocity of joint particle mass i.e(m

_{1}+m_{2}) would be zero.Therefore in center of mass frame of refrence ,particle which stick together will remain stationary.

**A) In labaoratory frame of refrence**

- Let m
_{1}and m_{2}be the two mass particle in a laboratory frame of refrence and m_{1}collide with m_{2}which is initailly is at rest.Let the velocity of mass m_{1}before collison be**u**and after the collison it moves with a velocity_{1}**v**and is delfected by the angle θ_{1}_{1}withs its incident direction and

m_{2}after the collision moves with the velocity**v**and it is deflected by an angle θ_{2}_{2}with its incident direction

- From law of conservation of linear momentum ,for components along x-axis

m_{1}u_{1}=m_{1}v_{1}cosθ_{1}+m_{2}v_{2}cosθ_{2}---(1)

For components along y-axis

0=m_{1}v_{1}sinθ_{1}-m_{2}v_{2}sinθ_{2}--(2)

And from law of conservation of energy

(1/2)m_{1}u_{1}^{2}=(1/2)m_{1}v_{1}^{2}+(1/2)m_{2}v_{2}^{2}---(3) - Analysing above equations we come to know that we have to find values of four unknown quantities v
_{1},v_{2},θ_{1},θ_{2}with the help of above three equations which is impossible as we need to have atleast four equations for finding out the values of four unknown quantities .Hence this problem can be solved in C frame of refrence

**B) In center of mass frame of refrence**

- Velocity of the center of mass wrt to L -frame of refrence is equation

- In C frame of refrence ,the center of mass remain stationary ,the velocity of mass m
_{1}before collision w.r.t C-frame of refrence is

**u**=_{1}^{'}**u**-_{1}**v**(5)_{cm}

And that of mass m_{2}is

**u**=0 -_{2}^{'}**v**= -_{cm}**v**(6)_{cm}

After the collision velocites of m_{1}and m_{2}in C-frame of refrence would be

**v**=_{1}^{'}**v**-_{1}**v**(7)_{cm}

**v**=_{2}^{'}**v**-_{2}**v**(8)_{cm}

Since center of mass remains stationary in C-frame of refrence hence total momentum would be zero.Therefore the momentum of both the massed would be equal and in opposite direction

So

m_{1}**u**= -m_{1}^{'}_{2}**u**--(9)_{2}^{'}

m_{1}**v**= -m_{1}^{'}_{2}**v**--(10)_{2}^{'}

The above equation can be proved like this

m_{1}**u**= m_{1}^{'}_{1}(**u**-_{1}**v**)_{cm}

and -m_{2}**u**= m_{2}^{'}_{2}**v**_{cm}

From the above equations

m_{1}**u**= -m_{1}^{'}_{2}**u**_{2}^{'}

Similarly we can prove that

m_{1}**v**= -m_{1}^{'}_{2}**v**_{2}^{'}

- It is clear from the above equations that after collisions the velocity of the particles i.e
**v**,_{1}^{'}**v**would be in opposite direction to each other and make same angles with the direction of the intial velocities of the particles as shown fig: (b)_{2}^{'}

- From equation (9) and (10)

**u**=(m_{2}^{'}_{1}/m_{2})**u**_{1}^{'}

and**v**=(m_{2}^{'}_{1}/m_{2})b>v_{1}^{'}

- From the law of conservation of energy in this frame of refrence

- From which |
**v**|=|_{1}^{'}**u**| and |_{1}^{'}**v**|=|_{2}^{'}**u**|._{2}^{'} - From these equations we came to know that magnitude of the velocities in C-frame of refrance , when the collision is elastic, does not change but their direction could change after collision.

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