A ball is thrown vertically downwards from a height of 20 m with an initial velocity v_{0}. It collides with the ground, loses 50% of its energy in a collision, and rebounds to the same height. The initial velocity v_{0 }is:

[Take, g = 10 ms^{-1}]

1. 14 ms^{-1}

2. 20 ms^{-1}

3. 28 ms^{-1}

4. 10 ms^{-1}

Subtopic: Work Energy Theorem |

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Two similar springs P and Q have spring constants k_{P} and k_{Q}, such that k_{P}> k_{Q}. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs W_{P }and W_{Q} are related as, in case (a) and case (b), respectively :

1. \(W_P=W_Q;W_P>W_Q\)

2. \(W_P=W_Q;W_P=W_Q\)

3. \(W_P>W_Q;W_P<W_Q\)

4. \(W_P<W_Q;W_P<W_Q\)

Subtopic: Work Energy Theorem |

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A block of mass 10 kg, moving in the x-direction with a constant speed of 10 ms^{-1} is subjected to a retarding force F = 0.1x J/m during its travel from x = 20 m to 30 m. Its final K.E. will be:

1. 475 J

2. 450 J

3. 275 J

4. 250 J

Subtopic: Work Energy Theorem |

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A particle of mass m is driven by a machine that delivers a constant power of k watts. If the particle starts from rest, the force on the particle at time t is:

1. $\sqrt{\frac{mk}{2}}{t}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

2. $\sqrt{mk}{t}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

3. $\sqrt{2mk}{t}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

4. $\frac{1}{2}$$\sqrt{mk}{t}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Subtopic: Power |

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Two particles of masses m_{1} and m_{2} move with initial velocities u_{1} and u_{2} respectively. On collision, one of the particles gets excited to a higher level, after absorbing energy E. If the final velocities of particles are v_{1} and v_{2}, then we must have:

1. ${\mathrm{m}}_{1}^{2}{\mathrm{u}}_{1}+{\mathrm{m}}_{2}^{2}{\mathrm{u}}_{2}-\mathrm{E}={\mathrm{m}}_{1}^{2}{\mathrm{v}}_{1}+{\mathrm{m}}_{2}^{2}{\mathrm{v}}_{2}$

2. $\frac{1}{2}{\mathrm{m}}_{1}{\mathrm{u}}_{1}^{2}+\frac{1}{2}{\mathrm{m}}_{2}{\mathrm{u}}_{2}^{2}=\frac{1}{2}{\mathrm{m}}_{1}{\mathrm{v}}_{1}^{2}+\frac{1}{2}{\mathrm{m}}_{2}{\mathrm{v}}_{2}^{2}$

3. $\frac{1}{2}{\mathrm{m}}_{1}{\mathrm{u}}_{1}^{2}+\frac{1}{2}{\mathrm{m}}_{2}{\mathrm{u}}_{2}^{2}-\mathrm{E}=\frac{1}{2}{\mathrm{m}}_{1}{\mathrm{v}}_{1}^{2}+\frac{1}{2}{\mathrm{m}}_{2}{\mathrm{v}}_{2}^{2}$

4. $\frac{1}{2}{\mathrm{m}}_{1}^{2}{\mathrm{u}}_{1}^{2}+\frac{1}{2}{\mathrm{m}}_{2}^{2}{\mathrm{u}}_{2}^{2}+\mathrm{E}=\frac{1}{2}{\mathrm{m}}_{1}^{2}{\mathrm{v}}_{1}^{2}+\frac{1}{2}{\mathrm{m}}_{2}^{2}{\mathrm{v}}_{2}^{2}$

Subtopic: Collisions |

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On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of the same mass M which is initially at rest. After the collision, the first block moves at an angle $\theta $ to its initial direction and has a speed $\frac{v}{3}$. The second block’s speed after the collision will be:

1. $\frac{2\sqrt{2}}{3}v$

2. $\frac{3}{4}v$

3. $\frac{3}{\sqrt{2}}v$

4. $\frac{\sqrt{3}}{2}v$

Subtopic: Collisions |

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A uniform force of \((3 \hat{i} + \hat{j})\) newton acts on a particle of mass 2 kg. Hence the particle is displaced from position \((2 \hat{i} + \hat{k})\) meter to position \((4 \hat{i} + 3 \hat{j} - \hat{k})\) meter. The work done by the force on the particle is:

1. 6 J

2. 13 J

3. 15 J

4. 9 J

Subtopic: Concept of Work |

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The potential energy of a particle in a force field is U=$\frac{A}{{r}^{2}}-\frac{B}{r}$ where A and B are positive constants and r is the distance of the particle from the center of the field. For stable equilibrium, the distance of the particle is:

1. B/A

2. B/2A

3. 2A/B

4. A/B

Subtopic: Potential Energy: Relation with Force |

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Two spheres A and B of masses ${\mathrm{m}}_{1}$ and ${\mathrm{m}}_{2}$, respectively, collide. A is at rest initially and B is moving with velocity v along the x-axis. After collision, B has a velocity $\frac{\mathrm{v}}{2}$ in a direction perpendicular to the original direction. The mass A moves after collision in the direction:

1. same as that of B.

2. opposite to that of B.

3. $\mathrm{\theta}={\mathrm{tan}}^{-1}\left(\frac{1}{2}\right)$ to the positive x-axis.

4. $\mathrm{\theta}={\mathrm{tan}}^{-1}\left(\frac{-1}{2}\right)$ to the positive x-axis

Subtopic: Collisions |

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A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a

constant magnitude ${P}_{0}$. The instantaneous velocity of this car is proportional to -

1. ${t}^{1/2}$

2. ${t}^{-1/2}$

3. $t/\sqrt{m}$

4. ${t}^{2}{P}_{0}$

Subtopic: Power |

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