Question

derive elastic collision in two direction

Answer 1

Hi,

Here is your answer,

Let us consider two objects A and B of mass 'm' m₁ and m₂ kept on the X-axis. Initially, the objects B is at rest and A moves toward with a speed u₁. If the collision is not head-on ( the force during the collision is not along the initial velocity ), the objects move along different lines. Let us imagine the object A moves with a velocity v₁ making an angle θ with the X-axis and the objects B moves with a velocity v₂ making an angle Φ with the same axis. Also, suppose v₁⁻ and v₂⁻ lie in X-Y plane. Using conservation of momentum in X and Y directions.

m₁u₁ = m₁v₁ cosθ + m₂v₂ cosΦ ----------- (1)

and 0 = m₁v₁ sinθ - m₂v₂ sinΦ --------------> (2)

If the collision is elastic, the final kinetic energy is equal to the initial velocity energy.

1/2 m₁u₁² = 1/2 m₁v₁² + 1/2 m₂v₂²

→ Now, we have four unknowns v₁ , v₂ , θ and Φ to describe the final motion whereas there are only three relations.

→ So, therefore he final motion depends on the angle between the line of force during and the direction of initial velocity. The momentum of each object must be individually conserved in the direction perpendicular to the force. The motion along the line of force may be treated as a 1-dimensional collision.

Hope it helps you !