Work, Energy and Power
- A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4 ms–1. It collides with a horizontal spring of force constant 200 Nm–1. The maximum compression produced in the spring will be :
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At maximum compression the solid cylinder will stop so loss in K.E. of cylinder = gain in P.E. of spring
⇒ 1 mv2 + 1 Iω2 = 1 kx2 2 2 2 ⇒ 1 mv2 + 1 mR2 
v 
2 = 1 kx2 2 2 2 R 2 ⇒ 3 mv2 = 1 kx2 4 2 ⇒ 3 × 4 × (4)2 = 1 × 200 x2 4 2 ⇒ 36 = x2 ⇒ x = 0.6 m 100
Correct Option: B
At maximum compression the solid cylinder will stop so loss in K.E. of cylinder = gain in P.E. of spring
⇒ 1 mv2 + 1 Iω2 = 1 kx2 2 2 2 ⇒ 1 mv2 + 1 mR2 v 2 = 1 kx2 2 2 2 R 2 ⇒ 3 mv2 = 1 kx2 4 2 ⇒ 3 × 4 × (4)2 = 1 × 200 x2 4 2 ⇒ 36 = x2 ⇒ x = 0.6 m 100
- A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of heaviest fragment in m/s will be
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Masses of the pieces are 1, 1, 3 kg. Hence
(1 × 2 l)2 + (1 × 2 l)2 = (3 × V)2
That is ,V = 7 √2 m / sCorrect Option: A
Masses of the pieces are 1, 1, 3 kg. Hence
(1 × 2 l)2 + (1 × 2 l)2 = (3 × V)2
That is ,V = 7 √2 m / s
- Two identical balls A and B moving with velocities +0.5 m/s and –0.3 m/s respectively, collide head on elastically. The velocities of the balls A and B after collision, will be, respectively
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When the identical balls collide head-on, their velocities are exchanged.
Correct Option: B
When the identical balls collide head-on, their velocities are exchanged.
- A shell is fired from a cannon, it explodes in mid air, its total
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When shell explodes in mid air its chemical energy is partly converted into mechanical energy, hence K.E. increases.
Correct Option: C
When shell explodes in mid air its chemical energy is partly converted into mechanical energy, hence K.E. increases.
- A body of mass m moving with velocity 3 km/h collides with a body of mass 2 m at rest. Now the coalesced mass starts to move with a velocity
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Applying law of conservation of momentum,
m1u1 + m2u2 = (m1 + m2)v
or, m1u1 = (m1 + m2)v (∵ u2 = 0)⇒ m (3 × 1000) = 3m(v) 3600 ⇒ v = 1000 m/s = 1 km / hr 3600 Correct Option: A
Applying law of conservation of momentum,
m1u1 + m2u2 = (m1 + m2)v
or, m1u1 = (m1 + m2)v (∵ u2 = 0)⇒ m (3 × 1000) = 3m(v) 3600 ⇒ v = 1000 m/s = 1 km / hr 3600
