Work, Energy and Power
- A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k = 50 N/m. The maximum compression of the spring would be
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1 mv2 = 1 kx2 2 2
⇒ mv2 = kx2
or 0.5 × (1.5)2 = 50 × x2
∴ x = 0.15 mCorrect Option: B
1 mv2 = 1 kx2 2 2
⇒ mv2 = kx2
or 0.5 × (1.5)2 = 50 × x2
∴ x = 0.15 m
- A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 ms–1. The kinetic energy of the other mass is
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From conservation of linear momentum m1v1 + m2v2 = 0
v2 = 
-m1 
v1 = -18 6 = - 9 ms-1 m2 12 K.E. = 1 m1v22 + 1 × 12 × 92 = 486 J 2 2
Correct Option: B
From conservation of linear momentum m1v1 + m2v2 = 0
v2 = -m1 v1 = -18 6 = - 9 ms-1 m2 12 K.E. = 1 m1v22 + 1 × 12 × 92 = 486 J 2 2
- An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, 1 kg first part moving with a velocity of 12 ms–1 and 2 kg second part moving with a velocity of 8 ms–1. If the third part flies off with a velocity of 4 ms–1, its mass would be :
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Let two parts of the rock move along x-axis and y-axis respectively.
If M and v be the mass and velocities of third part then
Mv cos θ = 12
Mv sin θ =16tan θ = 16 = 4 12 3 cos θ = 3 5
v = 4 m /sM = 12 v cos θ M = 12 × 5 = 60 = 5 kg 4 × 3 12 
Correct Option: D
Let two parts of the rock move along x-axis and y-axis respectively.
If M and v be the mass and velocities of third part then
Mv cos θ = 12
Mv sin θ =16tan θ = 16 = 4 12 3 cos θ = 3 5
v = 4 m /sM = 12 v cos θ M = 12 × 5 = 60 = 5 kg 4 × 3 12
- A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in m/s) after collision will be :
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Clearly v1 = 2 ms –1, v2 = 0
m1 = m (say), m2 = 2m
v1' = ?, v'2 = ?e = v1' - v2' ....(i) v2 - v1
By conservation of momentum, 2m = mv1' + 2mv2'... (ii)
From (i),0.5 = v2' - v1' 2
∴ v2' = 1 + v1'
From (ii), 2 = v1'+ 2 + 2 v1'
v1' = 0 and v2' = 1 ms–1Correct Option: A
Clearly v1 = 2 ms –1, v2 = 0
m1 = m (say), m2 = 2m
v1' = ?, v'2 = ?e = v1' - v2' ....(i) v2 - v1
By conservation of momentum, 2m = mv1' + 2mv2'... (ii)
From (i),0.5 = v2' - v1' 2
∴ v2' = 1 + v1'
From (ii), 2 = v1'+ 2 + 2 v1'
v1' = 0 and v2' = 1 ms–1
- A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is
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As the two masses stick together after collision, hence it is inelastic collision. Therefore, only momentum is conserved.
∴ mvî + 3m(2v)ĵ = (4m)v→v→ = v î + 6 vĵ 4 4 = v î + 3 vĵ 4 2 Correct Option: A
As the two masses stick together after collision, hence it is inelastic collision. Therefore, only momentum is conserved.
∴ mvî + 3m(2v)ĵ = (4m)v→v→ = v î + 6 vĵ 4 4 = v î + 3 vĵ 4 2
