The Question is from neet-physics work-energy-power cofficient-of-restitution, and it was created by the author VENKATESWARARAO

Question

Two identical balls collide head-on elastically.  If the initial velocity of one ball is $u$ and the other is at rest, what are their velocities after the collision?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [C]
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of Momentum and Kinetic Energy in Elastic Collisions
	**Related Topics**: Linear Momentum, Kinetic Energy, Elastic Collision
	**Prerequisites**: Understanding of momentum and kinetic energy, elastic collisions.
## SOLUTION PATHWAY
**Given Information**:
	* Two identical balls (same mass, say 'm')
	* Ball 1 initial velocity: $u_1 = u$
	* Ball 2 initial velocity: $u_2 = 0$
	* Elastic collision (both momentum and kinetic energy are conserved)
**Method & Steps**:
	1. Apply conservation of momentum:
		Reasoning: In a closed system, the total momentum before and after the collision remains the same.
		Formula/Rule: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ where $v_1$ and $v_2$ are the final velocities.
	2. Apply conservation of kinetic energy:
		Reasoning: In an elastic collision, the total kinetic energy before and after the collision remains the same.
		Formula/Rule: $\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$
	3. Substitute the given values and solve the equations:
		Working: Let m be the mass of each ball.  Momentum conservation: $mu + m(0) = mv_1 + mv_2$ which simplifies to $u = v_1 + v_2$. Kinetic energy conservation: $\frac{1}{2}mu^2 + \frac{1}{2}m(0)^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2$ which simplifies to $u^2 = v_1^2 + v_2^2$.  From the momentum equation, we have $v_1 = u - v_2$. Substituting this into the kinetic energy equation, we get $u^2 = (u-v_2)^2 + v_2^2 = u^2 - 2uv_2 + v_2^2 + v_2^2$. Simplifying: $0 = -2uv_2 + 2v_2^2$, so $2v_2(v_2 - u) = 0$.  $v_2 = 0$ or $v_2 = u$. $v_2$ cannot be 0 because that means there was no collision, so $v_2 = u$. Then, $v_1 = u - v_2 = u - u = 0$.
	4. Final velocities:
		Result: $v_1 = 0$ and $v_2 = u$
		Verification: The initial momentum was $mu$ and the final momentum is $m(0) + mu = mu$. The initial kinetic energy was $\frac{1}{2}mu^2$ and the final kinetic energy is $\frac{1}{2}m(0)^2 + \frac{1}{2}mu^2 = \frac{1}{2}mu^2$. Both momentum and kinetic energy are conserved.
 ## OPTION ANALYSIS
[A] :x: This would imply only partial transfer of momentum and energy.
[B] :x: This would imply no change in velocity, which is incorrect in a collision.
[C] ✓: This is the correct result derived from conservation of momentum and kinetic energy.
[D] :x: This would violate conservation laws.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting to use both conservation of momentum and kinetic energy in elastic collisions. [How to avoid]: Always apply both conservation laws for elastic collisions.
		2.[Error 2]: Incorrectly setting up the equations or making algebraic mistakes. [How to avoid]: Write down the equations carefully and double-check your algebra.
	**Quick Tips**:
		1.[Helpful shortcut]: In equal mass elastic head on collisions, the velocities are exchanged.
		2.[Memory aid]: Remember the conservation laws: momentum and kinetic energy are conserved in elastic collisions.

VENKATESWARARAO · Answer

$u/2$ and $u/2$

VENKATESWARARAO · Answer

$u$ and 0

VENKATESWARARAO · Answer

The correct Option is "0 and $u$"

VENKATESWARARAO · Answer

-$u$ and $2u$

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