A Disc of mass m moving on a horizontal surface with velocity V collides elastically identical stationary disc. The velocity of discs after collision are ... Plz help me to solve this question ..Thanks for ur help sir !

To solve the problem of two identical discs colliding elastically, we need to apply the principles of conservation of momentum and conservation of kinetic energy. Let's break this down step by step.

Understanding the Scenario

We have two discs, both with mass \( m \). One disc is moving with an initial velocity \( V \), while the other is stationary. After the collision, we want to find the velocities of both discs.

Key Principles to Apply

• Conservation of Momentum: The total momentum before the collision equals the total momentum after the collision.
• Conservation of Kinetic Energy: In an elastic collision, the total kinetic energy before the collision equals the total kinetic energy after the collision.

Setting Up the Equations

Let's denote the initial moving disc as Disc 1 and the stationary disc as Disc 2. The initial momentum and kinetic energy can be expressed as follows:

• Initial momentum: \( mV + 0 = mV \)
• Initial kinetic energy: \( \frac{1}{2}mV^2 + 0 = \frac{1}{2}mV^2 \)

After the collision, let the velocity of Disc 1 be \( V_1' \) and the velocity of Disc 2 be \( V_2' \). The equations for momentum and kinetic energy after the collision are:

• Final momentum: \( mV_1' + mV_2' = mV \)
• Final kinetic energy: \( \frac{1}{2}mV_1'^2 + \frac{1}{2}mV_2'^2 = \frac{1}{2}mV^2 \)

Solving the Equations

From the conservation of momentum, we can simplify the equation:

Dividing through by \( m \):

\( V_1' + V_2' = V \) (1)

From the conservation of kinetic energy, we can also simplify:

Dividing through by \( \frac{1}{2}m \):

\( V_1'^2 + V_2'^2 = V^2 \) (2)

Using the Equations Together

Now we have two equations to work with. We can use equation (1) to express \( V_1' \) in terms of \( V_2' \):

\( V_1' = V - V_2' \)

Substituting this into equation (2):

\( (V - V_2')^2 + V_2'^2 = V^2 \)

Expanding the first term:

\( V^2 - 2V V_2' + V_2'^2 + V_2'^2 = V^2 \)

Combining like terms gives:

\( V^2 - 2V V_2' + 2V_2'^2 = V^2 \)

Now, simplifying further:

\( -2V V_2' + 2V_2'^2 = 0 \)

Factoring out \( 2V_2' \):

\( 2V_2'(V_2' - V) = 0 \)

This gives us two solutions: \( V_2' = 0 \) (which we discard since it implies no movement) or \( V_2' = V \). Substituting \( V_2' = V \) back into equation (1):

\( V_1' + V = V \) implies \( V_1' = 0 \).

Final Velocities

After the elastic collision, the results are:

• Velocity of Disc 1 after collision, \( V_1' = 0 \)
• Velocity of Disc 2 after collision, \( V_2' = V \)

This means that the initially moving disc comes to a stop, and the stationary disc moves away with the same velocity that the first disc had before the collision. This outcome is typical in elastic collisions between identical objects where one is initially at rest.