Select Chapter Topics: A ball is thrown vertically downward 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, $$g=10~\mathrm{ms^{-2}}$$]
1. $$14$$ ms-1
2. $$20$$ ms-1
3. $$28$$ ms-1
4. $$10$$ ms-1  Subtopic:  Work Energy Theorem |
64%
From NCERT
NEET - 2015
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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 |
72%
From NCERT
NEET - 2015
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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 kinetic energy will be:
1. $$475$$ J
2. $$450$$ J
3. $$275$$ J
4. $$250$$ J  Subtopic:  Work Energy Theorem |
71%
From NCERT
NEET - 2015
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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. $$\sqrt{\frac{m k}{2}} t^{-1 / 2}$$
2. $$\sqrt{m k} t^{-1 / 2}$$
3. $$\sqrt{2 m k} t^{-1 / 2}$$
4. $$\frac{1}{2} \sqrt{m k} t^{-1 / 2}$$  Subtopic:  Power |
54%
From NCERT
NEET - 2015
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Two particles of masses m1 and m2 move with initial velocities u1 and u2 respectively. On collision, one of the particles gets excited to a higher level, after absorbing energy E. If the final velocities of particles are v1 and v2, 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 |
62%
From NCERT
NEET - 2015
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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 |
66%
From NCERT
NEET - 2015
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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 |
80%
From NCERT
AIPMT - 2013
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A particle with total energy $$E$$ is moving in a potential energy region $$U(x).$$ The motion of the particle is restricted to the region where:
1. $$U(x)<E$$
2. $$U(x)=0$$
3. $$U(x)\leq E$$
4. $$U(x)> E$$  Subtopic:  Conservation of Mechanical Energy |
From NCERT
NEET - 2013
Hints

One coolie takes $$1$$ minute to raise a suitcase through a height of $$2$$ m but the second coolie takes $$30$$ s to raise the same suitcase to the same height. The powers of two coolies are in the ratio:
1. $$1:3$$
2. $$2:1$$
3. $$3:1$$
4. $$1:2$$  Subtopic:  Power |
86%
From NCERT
NEET - 2013
Hints

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 |
77%
From NCERT
AIPMT - 2012
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Hints 