1. Two spheres A and B of masses m₁ and m₂  respectively collide. A is at rest initially and
   

   B is moving with velocity v along x-axis. After collision B has a velocity	v	in a direction perpendicular
   	2

   to the original direction. The mass A moves after collision in the direction.

1. 1. Same as that of B

   
   
   

   2. Opposite to that of B ​
      

   
   
   

   3. θ = tan^–1 (1/2) to the x-axis

   
   
   

   4. θ = tan–1 (–1/2) to the x-axis

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   u = 0 ​
   conservation of linear momentum along x-direction
   

   m2v = m1vx ⇒	m2v	= vx
   	m1

   
   along y-direction
   

   m2 ×	v	= m1vy ⇒ vy =	m2v
   	2		2m1

   
   Note : Let A moves in the direction, which makes an angle θ with initial direction i.e.
   

   tan θ =	vy	=		m1v
   				2m1
   	vx			m2v
   				2m1

   
   

   tan θ =	1
   	2

   
   

   θ = tan-1		1		to the x-axis.
   		2

   
   

   Correct Option: C

   
   u = 0 ​
   conservation of linear momentum along x-direction
   

   m2v = m1vx ⇒	m2v	= vx
   	m1

   
   along y-direction
   

   m2 ×	v	= m1vy ⇒ vy =	m2v
   	2		2m1

   
   Note : Let A moves in the direction, which makes an angle θ with initial direction i.e.
   

   tan θ =	vy	=		m1v
   				2m1
   	vx			m2v
   				2m1

   
   

   tan θ =	1
   	2

   
   

   θ = tan-1		1		to the x-axis.
   		2

   
   

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2. 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. 1. 	3	v
      	4

   
   
   

   2. 	3	v
      	√2

   
   
   

   3. 	√3	v
      	2

   
   
   

   4. 	2√2	v
      	3

   
   
   

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1. View Hint View Answer Discuss in Forum

   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

   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

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3. An engine pumps water through a hose pipe.Water passes through the pipe and leaves it with a velocity of 2 m/s. The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine?
   ​​​

1. 1. 400 W

   
   
   

   2. 200 W

   
   
   

   3. 100 W

   
   
   

   4. 800 W

   
   
   

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1. View Hint View Answer Discuss in Forum

   ​Amount of water flowing per second from the pipe
   

   =	m	=	m	.	l	=		m		v
   	time		l		t			l

   
   Power = K.E. of water flowing per second
   

   =	1			m		v . v2
   	2			l

   
   

   =	1			m		v3
   	2			l

   
   

   =	1	× 100 × 8 = 400 W
   	2

   
   

   Correct Option: D

   ​Amount of water flowing per second from the pipe
   

   =	m	=	m	.	l	=		m		v
   	time		l		t			l

   
   Power = K.E. of water flowing per second
   

   =	1			m		v . v2
   	2			l

   
   

   =	1			m		v3
   	2			l

   
   

   =	1	× 100 × 8 = 400 W
   	2

   
   

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4. A particle of  mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10^–4 J by the end of the second revolution after the beginning of the motion ?​​​
   

1. 1. ​0.1 m/s²

   
   
   

   2. 0.15 m/s²

   
   
   

   3. 0.18 m/s²

   
   
   

   4. 0.2 m/s²

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given : Mass of particle, M = 10g =	10	kg
   	1000

   
   radius of circle R = 6.4 cm ​
   Kinetic energy E of particle = 8 × 10^–4J ​
   acceleration a_t = ?
   

   	1	mv2 = E ⇒	1			10		v2 = 8 × 10–4
   	2		2			1000

   
   ⇒ v² = 16 × 10^–2
   ⇒ v = 4 × 10^–1 = 0.4 m / s
   Now, using ​v² = u² + 2a_ts ​(s = 4πR)
   

   (0.4)2 = 02 + 2at		4 ×	22	×	6.4
   			7		100

   
   

   ⇒ at = (0.4)2 ×	7 × 100	= 0.1 m / s2
   	8 × 22 × 6.4

   
   

   Correct Option: A

   Given : Mass of particle, M = 10g =	10	kg
   	1000

   
   radius of circle R = 6.4 cm ​
   Kinetic energy E of particle = 8 × 10^–4J ​
   acceleration a_t = ?
   

   	1	mv2 = E ⇒	1			10		v2 = 8 × 10–4
   	2		2			1000

   
   ⇒ v² = 16 × 10^–2
   ⇒ v = 4 × 10^–1 = 0.4 m / s
   Now, using ​v² = u² + 2a_ts ​(s = 4πR)
   

   (0.4)2 = 02 + 2at		4 ×	22	×	6.4
   			7		100

   
   

   ⇒ at = (0.4)2 ×	7 × 100	= 0.1 m / s2
   	8 × 22 × 6.4

   
   

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5. A body of  mass 3 kg is under a constant force which causes a displacement s in metres in

   it,given by the relation s =	1	t2 , where t is in seconds. Work done by the force in 2 seconds is
   	3

1. 1. 	3	J
      	8

   
   
   

   2. 	8	J
      	3

   
   
   

   3. 	19	J
      	5

   
   
   

   4. 	5	J
      	19

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Acceleration =	d2s	=	2	m / s2
   	dt2		3

   
   

   Force acting on the body = 3 ×	2	= 2 newton
   	3

   
   

   Displacement in 2 secs =	1	× 2 × 2 =	4	m
   	3		3

   
   

   Work done = 2 ×	4	=	8	J
   	3		3

   Correct Option: B

   Acceleration =	d2s	=	2	m / s2
   	dt2		3

   
   

   Force acting on the body = 3 ×	2	= 2 newton
   	3

   
   

   Displacement in 2 secs =	1	× 2 × 2 =	4	m
   	3		3

   
   

   Work done = 2 ×	4	=	8	J
   	3		3

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