What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

1. $\sqrt{2\mathrm{gR}}$

2. $\sqrt{3gR}$

3. $\sqrt{5gR}$

4. $\sqrt{gR}$
 

A block of mass \(10\) kg, moving in the \(x\text-\)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:

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. $\sqrt{\left(\frac{mk}{2}\right)}{t}^{-1/2}$
 

2. $\sqrt{\left(mk\right)}{t}^{-1/2}$
 

3. $\sqrt{\left(2mk\right)}{t}^{-1/2}$
 

4. $\frac{1}{2}\sqrt{\left(mk\right)}{t}^{-1/2}$

Two particles of masses m₁,m₂ move with initial velocities u₁ and u₂. On collision, one of the particles get excited to higher level, after absorbing energy $\epsilon$. If final velocities of particles be v₁ and v₂, then we must have

1. m₁²u₁+m₂²u₂-$\epsilon$=m₁²v₁+m₂²v₂

2. $\frac{1}{2}$m₁u₁²+$\frac{1}{2}$m₂u₂=$\frac{1}{2}$m₁v₁²+$\frac{1}{2}$m₂v₂²-$\epsilon$

3. $\frac{1}{2}$m₁u₁²+$\frac{1}{2}$m₂u₂²-$\epsilon$=$\frac{1}{2}$m₁v₁²+$\frac{1}{2}$m₂v₂²

4. $\frac{1}{2}$m₁²u₁²+$\frac{1}{2}$m₂²u₂²+$\epsilon$=$\frac{1}{2}$m₁²v₁²+$\frac{1}{2}$m₂²v₂²
 

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 \(g = 10~\text{m/s}^2\))
1. \(14~\text{m/s}\)
2. \(20~\text{m/s}\)
3. \(28~\text{m/s}\)
4. \(10~\text{m/s}\)

On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle θ to its initial direction and has a speed v/3. The second block's speed after the collision is:

1. 2√2v/3
 

2. 3v/4
 

3. 3v/√2
 

4. √3v/2

A uniform force of (3i + j) N acts on a particle of mass 2 kg. Hence the particle is displaced from position (2i+k) m to position (4i+3j-k) m. The work done by the force on the particle is-

1. 9J
 

2. 6J
 

3. 13J
 

4. 15J

The potential energy of a system increases if work is done

(1) by the system against a conservative force 

(2) by the system against a nonconservative force

(3) upon the system by a conservative force

(4) upon the system by a nonconservative force

Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 m is

(a) 21 J                                                  (b) 26 J

(c) 13 J                                                   (d) 18 J