Physics, asked by vaiswarya733, 4 months ago

derive an expression for final velocities of two bodies which undergo elastic in one dimension ?

Answers

Answered by Anonymous
0

here's ur answer dude

Let two bodies of masses m

1

and m

2

moving with velocities u

1

and u

2

along the same straight line.

And consider the two bodies collide and after collision v

1

and v

2

be the velocities of two masses.

Before collision

Momentum of mass m

1

=m

1

u

1

Momentum of mass m

2

=m

2

u

2

Total momentum before collision

p

1

=m

1

u

1

+m

2

u

2

Kinetic energy of mass m

1

=

2

1

m

1

u

1

2

Kinetic energy of mass m

2

=

2

1

m

2

u

2

2

Thus,

Total kinetic energy before collision is

K.E=

2

1

m

1

u

1

2

+

2

1

m

2

u

2

2

After collision

Momentum of mass m

1

=m

1

v

1

Momentum of mass m

2

=m

2

v

2

Total momentum before collision

P

f

=m

1

v

1

+m

2

v

2

Kinetic energy of mass m

1

=

2

1

m

1

v

1

2

Kinetic energy of mass m

2

=

2

1

m

2

v

2

2

Total kinetic energy after collision

K

f

=

2

1

m

1

v

1

2

+

2

1

m

2

v

2

2

So, according to the law of conservation of momentum

m

1

u

1

+m

2

u

2

=m

1

v

1

+m

2

v

2

⇒m

1

(u

1

−v

1

)=m

2

(v

2

−u

2

)

............(1)

And according to the law of conservation of kinetic energy

2

1

m

1

u

1

2

+

2

1

m

2

u

2

2

=

2

1

m

1

v

1

2

+

2

1

m

2

v

2

2

⇒m

1

(u

1

2

−v

1

2

)=m

2

(v

2

2

−v

1

2

)

.............(2)

Now dividing the equation

(u

1

+v

1

)=(v

2

+u

2

)

⇒u

1

−u

2

=v

2

−v

1

Therefore, this is, relative velocity of approach is equal to relative velocity of separation.

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