46. The co-efficient of restitution e for a perfectly elastic collision is​​

1. 1. 1
      

   
   
   

   2. ​0

   
   
   

   3. ∞
      

   
   
   

   4. -1

   
   
   

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1. View Hint View Answer Discuss in Forum

   e = | v₁ - v₂ | / | u₁ - u₂ | which is 1 for a perfectly elastic collision.
   

   Correct Option: A

   e = | v₁ - v₂ | / | u₁ - u₂ | which is 1 for a perfectly elastic collision.
   

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47. An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 ms^–1 and the second part of mass 2 kg moves with speed 8 ms^–1. If the third part flies off with speed 4 ms^–1 then its mass is
    

1. 1. 5 kg

   
   
   

   2. 7 kg

   
   
   

   3. 17 kg

   
   
   

   4. 3 kg

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   P_resultant = √12² + 16²
   = √144 + 256 = 20
   m₃v₃ = 20 (momentum of third part)
   

   m3 =	20	= 5 kg
   	4

   
   

   Correct Option: A

   
   P_resultant = √12² + 16²
   = √144 + 256 = 20
   m₃v₃ = 20 (momentum of third part)
   

   m3 =	20	= 5 kg
   	4

   
   

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48. A body of mass (4m) is lying in x-y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds (v). The total kinetic energy generated due to explosion is :​​​
    

1. 1. mv²

   
   
   

   2. 	3	mv2
      	2

   
   
   

   3. ​2 mv²

   
   
   

   4. ​4 mv²

   
   
   

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1. View Hint View Answer Discuss in Forum

   By conservation of linear momentum ​​
   

   2mv1 = √2 mv ⇒ v1 =	v
   	√2

   
   
   As two masses of each of mass m move perpendicular to each other. ​
   

   Total KE generated =	1	mv2 +	1	mv2 +	1	(2m)v12
   	2		2		2

   
   

   = mv2 +	mv2	=	3	mv2
   	2		2

   Correct Option: B

   By conservation of linear momentum ​​
   

   2mv1 = √2 mv ⇒ v1 =	v
   	√2

   
   
   As two masses of each of mass m move perpendicular to each other. ​
   

   Total KE generated =	1	mv2 +	1	mv2 +	1	(2m)v12
   	2		2		2

   
   

   = mv2 +	mv2	=	3	mv2
   	2		2

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49. A ball is thrown vertically downwards from a height of 20 m with an initial velocity v₀. It collides with the ground loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v₀ is :​
    [2015 RS] ​(Take g = 10 ms^–2)

1. 1. 20 ms^–1

   
   
   

   2. 28 ms^–1

   
   
   

   3. ​10 ms^–1

   
   
   

   4. 14 ms^–1

   
   
   

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1. View Hint View Answer Discuss in Forum

   When ball collides with the ground it loses its 50% of energy
   

   ∴	K.Ef	=	1	⇒	1	m1Vf2		=	1
   					2
   	K.Ei		2		1	m2Vi2			2
   					2

   
   
   

   or	Vf	=	1
   	Vi		√2

   
   

   or	√2gh	=	1
   	√V0² + 2gh		√2

   
   or, 4gh = V₀² + 2gh
   ​∴ V₀ = 20 ms^–1
   

   Correct Option: A

   When ball collides with the ground it loses its 50% of energy
   

   ∴	K.Ef	=	1	⇒	1	m1Vf2		=	1
   					2
   	K.Ei		2		1	m2Vi2			2
   					2

   
   
   

   or	Vf	=	1
   	Vi		√2

   
   

   or	√2gh	=	1
   	√V0² + 2gh		√2

   
   or, 4gh = V₀² + 2gh
   ​∴ V₀ = 20 ms^–1
   

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50. ​Two particles A and B, move with constant velocities v^→₁ and v^→₂. At the initial moment their position vectors are r^→₁ and r^→₂ respectively. The condition for particles A and B for their collision is :​​​
    

1. 1. r^→₁ . v^→₁ = r^→₂ . v^→₂

   
   
   

   2. r^→₁ × v^→₁ = r^→₂ × v^→₂

   
   
   

   3. r^→₁ - r^→₂ = v^→₁ - v^→₂

   
   
   

   4. 	r1→ - r2→	=	v2→ - v1→
      	| r1→ - r2→ |		| v2→ - v1→ |

   
   
   

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1. View Hint View Answer Discuss in Forum

   For collision V_B/A^→ should be along
   B ≻ A(r_B/A^→)
   

   So ,	V2→ - V1→	=	r1→ - r2→
   	| V2 - V1 |		| r1 - r2 |

   
   
   

   Correct Option: D

   For collision V_B/A^→ should be along
   B ≻ A(r_B/A^→)
   

   So ,	V2→ - V1→	=	r1→ - r2→
   	| V2 - V1 |		| r1 - r2 |

   
   
   

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