1. A bullet spins as the shot is fired from a gun. For this purpose, two helical slots as shown in the figure are cut in the barrel. Projections A and B on the bullet engage in each of the slots.
   
   Helical slots are such that one turn of helix is completed over a distahce of 0.5 m. If velocity of bullet when it exits the barrel is 20 m/s, its spinning speed in rad/s is ______.

1. 1. 255.4 rad/sec

   
   
   

   2. 251.3 rad/sec

   
   
   

   3. 252.2 rad/sec

   
   
   

   4. 241.3 rad/sec

   
   
   

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1. View Hint View Answer Discuss in Forum

   Time taken force revolution =	0.5	= 0.025 sec
   	20

   
   

   The spinning speed is	2π	rad/sec = 251.3 rad/sec
   	0.02s

   Correct Option: B

   Time taken force revolution =	0.5	= 0.025 sec
   	20

   
   

   The spinning speed is	2π	rad/sec = 251.3 rad/sec
   	0.02s

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2. A mass m₁ of 100 kg travelling with a uniform velocity of 5 m/s along a line collides with a stationary mass m₂ of 1000 kg. After the collision, both the masses travel together with the same velocity. The coefficient of restitution is

1. 1. 0.6

   
   
   

   2. 0.1

   
   
   

   3. 0.01

   
   
   

   4. 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Coefficientof restitution =	Re lativespeedafter collision	= 0
   	Relativespeed beforecollision

   Correct Option: D

   Coefficientof restitution =	Re lativespeedafter collision	= 0
   	Relativespeed beforecollision

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3. A single degree of freedom system having mass 1 kg and stiffness 10 kN/m initially at rest is subjected to an impulse force of magnitude 5 kN for 10^–4 seconds. The amplitude in mm of the resulting free vibration is

1. 1. 0.5

   
   
   

   2. 1.0

   
   
   

   3. 5.0

   
   
   

   4. 10.0

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   At t = 0, mass is at rest, F = 5 kN for 10^–4 seconds & amplitude = x
   

   F = m ×	dv
   	dt

   
   F.^t₂∫^t₁ dt = m ^v₂∫^v₁dv
   F(t₂ – t₁) = m × (v₂ – v₁)
   5 × 10^–3 × 10_-4 = 1 × (v₂ – 0)
   0.5 m/s = v₂
   

   1	mv²2 =	1	kx²
   2		2

   
   

   ⇒ x = √	mv²2	= √	1 × 0.5 × 0.5	= 5 mm
   	k		10 × 1000

   Correct Option: C

   
   At t = 0, mass is at rest, F = 5 kN for 10^–4 seconds & amplitude = x
   

   F = m ×	dv
   	dt

   
   F.^t₂∫^t₁ dt = m ^v₂∫^v₁dv
   F(t₂ – t₁) = m × (v₂ – v₁)
   5 × 10^–3 × 10_-4 = 1 × (v₂ – 0)
   0.5 m/s = v₂
   

   1	mv²2 =	1	kx²
   2		2

   
   

   ⇒ x = √	mv²2	= √	1 × 0.5 × 0.5	= 5 mm
   	k		10 × 1000

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4. The coefficient of restitution of a perfectly plastic impact is

1. 1. 0

   
   
   

   2. 1

   
   
   

   3. 2

   
   
   

   4. ∞

   
   
   

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1. View Hint View Answer Discuss in Forum

   For perfectly plastic impact,
   

   Coefficient of restitution =	Relative velocity of separation	= 0
   	Relative velocity of approach

   Correct Option: A

   For perfectly plastic impact,
   

   Coefficient of restitution =	Relative velocity of separation	= 0
   	Relative velocity of approach

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5. A body of mass (M) 10 kg is initially stationary on a 45° inclined plane as shown in figure. The coefficient of dynamic friction between the body and the plane is 0.5. The body slides down the plane and attains a velocity of 20 m/s. The distance travelled (in meter) by the body along the plane is ________.
   

1. 1. 58.8 m.

   
   
   

   2. 56.8 m.

   
   
   

   3. 57.8 m.

   
   
   

   4. 58 m.

   
   
   

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1. View Hint View Answer Discuss in Forum

   FBD of block
   Net force = ma
   mg sin 45 – μmg cos 45 = ma
   
   a = 3.46 m/s²
   ⇒ V² – u² = 2as
   

   ⇒ S =	V²	=	20²	= 57.8 m.
   	2a		2 × 3.46

   Correct Option: C

   FBD of block
   Net force = ma
   mg sin 45 – μmg cos 45 = ma
   
   a = 3.46 m/s²
   ⇒ V² – u² = 2as
   

   ⇒ S =	V²	=	20²	= 57.8 m.
   	2a		2 × 3.46

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