Mechanical Engineering - Engineering Mechanics - Discussion

Velocity of the first particle of same mass after collision gets 0 and vel of sec particle after collision assumes the velocity of the first particle before collision( ie. V2 =U1, V1=0, U2=0,)

For a perfect elastic body property.

We know co-efficient of restitution,

e=(u1-v1)/(v2-u2)
Where u1,u2 are the initial velocities and v1,v2 are the final velocities.
For perfectly elastic material e=1,
And u1=0, u2= v and v1=v2,
So the eq becomes,
1=(0-v1)/(v2-v).
or, v2-v=-v1.
or, 2v1=v.
or, v1=v2=v/2.

It is newtons second law.

Let the initial velocity of first mass be u.

Final velocities attained by first and second masses be v and w respectively.

From conservation of momentum, mu = mv + mw,

We can have w = u - v ----->(1).

From conservation of kinetic energy,

1/2 mu2 = 1/2 mv2 + 1/2 mw2,

u2 = v2 + w2 ----->(2).

(1) in (2),

u2 = v2 + (u - v)2.

Simplifying gives 2 uv = 2 v2.

v = u.

In (1), w = 2u.

After collision ball 1 return with the same velocity and other attain a twice velocity.

Let the initial velocity of first mass be u.

Final velocities attained by first and second masses be v and w respectively.

From conservation of momentum, mu = mv + mw,

We can have w = u - v ----->(1).

From conservation of kinetic energy,

1/2 mu2 = 1/2 mv2 + 1/2 mw2,

u2 = v2 + w2 ----->(2).

(1) in (2),

u2 = v2 + (u - v)2.

Simplifying gives 2 uv = 2 v2.

v = u.

In (1), w = 2u.

After collision ball 1 return with the same velocity and other attain a twice velocity.

After impact, the ball 1 will come into rest and 2 will move with velocity we.

What you guys are saying that happens only if the collision is perfectly plastic?

For perfectly elastic body the velocity of each body is v/2.

mv1 + mv2 = mv'1+mv'2.
m is equal.

v1 + 0 = 2v' because v1 = 0 and v'1 = v'2.
v' = v/2.

According to the conservation of momentum.

The 1st ball will come to rest and the second ball will move with a velocity V.

mv1 + mv2 = mv1' + mv2'

Where m = mass (same for both the balls in this case)

v1 = v (velocity of moving ball, given)
v2 = 0 (velocity of stationary ball, given)

=> mv + m(0) = mv' + mv'.

mv = m(2v').

So, v1' = v2' = v/2.