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Principle of Conservation of EnergyTotal energy of an isolated system always remains constant. Since, the universe as a whole may be viewed as an isolated system, total energy of the universe is constant. If one part of the universe loses energy, then other part must gain an equal amount of energy. The principle of conservation of energy cannot be proved as such. However, no violation of this principle has been observed.In the given curved road, if particle is released from A,
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Principle of Conservation of EnergyTotal energy of an isolated system always remains constant. Since, the universe as a whole may be viewed as an isolated system, total energy of the universe is constant. If one part of the universe loses energy, then other part must gain an equal amount of energy. The principle of conservation of energy cannot be proved as such. However, no violation of this principle has been observed.`U` is the potential energy, K is the kinetic energy and E is the mechanical energy. Which of the following is not possible for a stable
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Principle of Conservation of EnergyTotal energy of an isolated system always remains constant. Since, the universe as a whole may be viewed as an isolated system, total energy of the universe is constant. If one part of the universe loses energy, then other part must gain an equal amount of energy. The principle of conservation of energy cannot be proved as such. However, no violation of this principle has been observed.A body of mass 5 kg is thrown vertically up with a kinetic energy of `490J`. The height at which the kinetic energy of the body becomes half of the
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A system of particles is called a rigid body,
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The centre of mass of a system of particles does not depend
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In pure rotation, all particles of the
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For n particles in a space, the suitable expression for the `x`-coordinate of the centre of mass of a system
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Two bodies of masses `1 kg and 2 kg` are lying on `x-y` plane at `(1, 2) and (- 1, 3)` respectively. What are the coordinates of centre of
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Three identical spheres of mass M each are placed at the corners of an equilateral triangle of side 2 m. Taking one of the corner as the origin, the position vector of the centre of mass
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Centre of mass of the given system of particles will be
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Two particles of equal masses have velocities `v_1 = 4 hati ms^(-1) and v_2 = 4 hatj ms^(-1)` . First particle has an acceleration `a_1 = (2hati + 2hatj) ms^(-2)`, while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path
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The centre of mass of three particles of masses `1 kg, 2 kg and 3 kg` is at `(3, 3, 3)` with reference to a fixed coordinate system. Where should a fourth particle of mass 4 kg be placed, so that the centre of mass of the system of all particles shifts to a point `(1, 1,
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A ball kept in a closed box moves in the box making collisions with the walls. The box is kept on a smooth surface. The velocity of the centre of
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A force `F` is applied on a single particle Pas shown in the figure. Here, `r` is the position vector of the particle. The value of torque `tau`
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A force `F=5hati + 2hatj - 5hatk` acts on a particle whose position vector is `r =hati - 2hatj + hatk`. What is the torque about the origin
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ABC is an equilateral triangle with `O` as its centre. `F_1, F_2 and F_3` represent three forces acting along the sides `AB, BC and AC,` respectively. If the total torque about `O` is zero, then the magnitude of `F_3`
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The angular momentum L of a single particle can be represented
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Newton’s second law for rotational motion of a system of particle can be represented as (L for a system of
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A particle of mass `m` moves in the `xy`-plane with a velocity `v` along the straight line AB. If the angular momentum of the particle with respect to origin `O` is `L_A`, when it is at A and `L_B` when it is at B,
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A point mass `m` is attached to a massless string whose other end is fixed at `P` as shown in figure. The mass is undergoing circular motion in `xy`-plane with centre `O` and constant angular speed `Omega`. If the angular momentum of the system, calculated about `O and P` be `L_O and L_P` respectively,
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