The Question is from neet-physics work-energy-power elastic-collisions, and it was created by the author VENKATESWARARAO

Question

Two identical smooth spheres collide obliquely.  Sphere A initially moves with speed $v$ at an angle $	heta$ to the line of centers, while sphere B is initially at rest. If the coefficient of restitution is $e$, what is the speed of sphere B after the collision?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option D
## CONCEPT FRAMEWORK
	**Primary Concept**: Conservation of Linear Momentum and Coefficient of Restitution
	**Related Topics**: Oblique Collisions, Impulse
	**Prerequisites**: Vector components, Collision theory
## SOLUTION PATHWAY
**Given Information**:
	*   Masses of spheres are identical (let's call it $m$)
	*   Initial velocity of A: $v$ at angle $	heta$ to the line of centers
	*   Initial velocity of B: 0
	*   Coefficient of restitution: $e$
	*   Collision is oblique
**Method & Steps**:
	1. Resolve the initial velocity of sphere A into components along the line of centers (x-direction) and perpendicular to it (y-direction).
		Reasoning: We need to analyze the motion along the line of impact separately from the motion perpendicular to it. The y-component of velocity will remain unchanged during the collision.
		Formula/Rule: $v_{Ax} = v \cos(	heta)$, $v_{Ay} = v \sin(	heta)$
	2. Apply the conservation of linear momentum and the definition of the coefficient of restitution along the line of centers (x-direction).
		Reasoning: Momentum is conserved in the x direction, and the coefficient of restitution relates the relative velocities after and before the collision in the x direction.
		Formula/Rule: Momentum Conservation: $m v_{Ax} + m (0) = m v_{A'x} + m v_{B'x}$  =>  $v \cos(	heta) = v_{A'x} + v_{B'x}$; Coefficient of Restitution: $e = \frac{v_{B'x} - v_{A'x}}{v_{Ax} - 0}$ => $e v \cos(	heta) = v_{B'x} - v_{A'x}$
	3. Solve the two equations obtained in step 2 simultaneously for $v_{B'x}$.
		Working: Adding the equations, we have $v \cos(	heta)(1+e) = 2 v_{B'x}$ => $v_{B'x} = \frac{v(1+e)}{2} \cos(	heta)$.
		Key Point: The y-component of velocity for sphere B remains zero since it was initially at rest and the impulse acts along the x-direction. $v_{By} = 0$
	4. Determine the speed of sphere B after the collision.
		Result: The speed of sphere B after the collision is $v_{B'} = \sqrt{v_{B'x}^2 + v_{B'y}^2} = \sqrt{(\frac{v(1+e)}{2} \cos(	heta))^2 + 0^2}$ = $\frac{v(1+e)}{2} \cos(	heta)$
		Verification: The units are consistent (speed), and the result makes sense in the context of the problem.
 ## OPTION ANALYSIS
[A] :x: This option misses the factor of 1/2.
[B] :x: This option uses $\sin(	heta)$ instead of $\cos(	heta)$ and is also missing a factor of 1/2.
[C] :x: This option has a (1-e) term instead of (1+e).
[D] ✓: This is the correct speed of sphere B after the collision.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly applying the coefficient of restitution: Make sure to use the correct relative velocities (final - initial) in the numerator and denominator.
		2. Not resolving velocities into components: Oblique collisions require analyzing motion along and perpendicular to the line of impact separately.
	**Quick Tips**:
		1. Remember that the velocity component perpendicular to the line of centers remains unchanged for each sphere.
		2. When solving collision problems, always draw a diagram with the velocities resolved into components.

VENKATESWARARAO · Answer

$v(1+e) \cos(	heta)$

VENKATESWARARAO · Answer

$\frac{v(1+e)}{2} \sin(	heta)$

VENKATESWARARAO · Answer

$\frac{v(1-e)}{2} \cos(	heta)$

VENKATESWARARAO · Answer

The correct Option is "$\frac{v(1+e)}{2} \cos(	heta)$"

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