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

(a)m₁²u₁+m₂²u₂-$\epsilon$=m₁²v₁+m₂²v₂

(b)$\frac{1}{2}$m₁u₁²+$\frac{1}{2}$m₂u₂=$\frac{1}{2}$m₁v₁²+$\frac{1}{2}$m₂v₂²-$\epsilon$

(c)$\frac{1}{2}$m₁u₁²+$\frac{1}{2}$m₂u₂²-$\epsilon$=$\frac{1}{2}$m₁v₁²+$\frac{1}{2}$m₂v₂²

(d)$\frac{1}{2}$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_o. It collides with the ground, loses 50% of its energy in a collision and rebounds to the same height. The initial velocity v_o is: (Take g = 10 ms^-2)

1. 14 ms^-1
2. 20 ms^-1
3. 28 ms^-1
4. 10 ms^-1

Two particles A and B. move with constant velocities v₁ and v₂. At the initial moment, their position vectors are $r_1$ and $r_2$ respectively. The condition for particles A and B for their collision is-


(1) $\frac{r_1-r_2}{\left|r_1-r_2\right|}=\frac{v_2-v_1}{\left|v_2-v_1\right|}$

(2) $r_1·v_1=r_2·v_2$

(3) _{$r_1\times v_1=r_2\times v_2$}

(4) r₁-r₂=v₁-v₂

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

The force F acting on a particle of mass m is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8 s is:
     
1. 24 Ns

2. 20 Ns

3. 12 Ns

4. 6 Ns

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