On a frictionless surface a block of mass M moving at speed

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On a frictionless surface a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle θ to its initial direction
and has a speed v . The second block's speed after the collision is :
3

1. 3 v
  4
2. 3 v
  √2
3. √3 v
  2
4. 2√2 v
  3

Correct Option: D

Here, M₁ = M₂ and u₂ = 0

u₁ = V , V₁ =	V	; V₂ = ?
	3

From figure, along x-axis, M₁u₁ + M₂u₂ = M₁V₁ cosθ + M₂V₂ cosφ ...(i)
Along y-axis 0 = M₁V₁ sinθ – M₂V_s sinφ ...(ii)
By law of conservation of kinetic energy

	1	M₁u₁² +	1	M₂u₂² =	1	M₁V₁² +	1	M₂V₂² ....(iii)
	2		2		2		2

Putting M₁ = M₂ and u₂ = 0 in equation (i), (ii) and (iii) we get

θ + Φ =	π	= 90°
	2

and u₁² = V₁² + V₂²

V² =			V		²	+ V₂²		∵ u₁ = V and V₁ =	V
			3						3

or , V² -			V		²	= V₂²
			3

V² -	V²	= V₂²
	9

or , V₂² =	8	V² ⇒ V₂ =	2 √2	V
	9		3

and has a speed	v	. The second block's speed after the collision is :
	3

	√3	v
	2

	2√2	v
	3