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 20m with an initial velocity v_o . It collides with the ground, loses 50% of it's energy in collision and rebounds to the same height. The initial velocity v_o is (Take, g = 10 ms^-2)

(1) 14 ms^-1

(2) 20ms^-1

(3) 28ms^-1

(4) 10ms^-1

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 particle from the centre of the field. For stable equilibrium, the distance of the particle is 

(a) B/2A                                     (b)2A/B

(c)A/B                                        (d)B/A

A car of mass m starts from rest and  accelerates so that the instantaneous power delivered to the car has a constant magnitude P_o. The instantaneous velocity of this car is proportional to

1. t₂P₀

2. t^1/2

3. t^–1/2

4. t / √m

The potential energy of a system increases if the 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

A ball moving with velocity $2 m{s}^{-1}$ collides head on with another stationery ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in $m{s}^{-1}$) after collision will be 

(a)0,1                                                (b)1,1

(c)1,0.5                                             (d)0,2 

A particle of mass M starting from rest undergoes uniform acceleration. If the speed acquired in time T is v, the power delivered to the particle is 

(1) $\frac{{\mathrm{Mv}}^2}{\mathrm{T}}$                                 

(2) $\frac{1}{2}\frac{{\mathrm{Mv}}^2}{{\mathrm{T}}^2}$

(3) $\frac{{\mathrm{Mv}}^2}{{\mathrm{T}}^2}$                                   

(4) $\frac{1}{2}\frac{{\mathrm{Mv}}^2}{\mathrm{T}}$

A body of mass 1 kg is thrown upwards with a velocity of $20$ ${\mathrm{ms}}^{-1}.$ It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? $\left(\mathrm{g}=10\right)$ ${\mathrm{ms}}^{-2}$

1. 20 J

2. 30 J

3. 40 J

4. 10 J

The points of maximum and minimum attraction in the curve between potential energy (U) and distance (r) of a diatomic molecules are respectively -

(a) Sand R

(b) T and S

(c) R and S

(d) S and T

K is the force constant of a spring. The work done in increasing its extension from $l_1$ to $l_2$ will be

(1) $K\left(l_2-l_1\right)$                             

(2) $\frac{K}{2}\left(l_2+l_1\right)$

(3) $K\left(l_2^2-l_1^2\right)$                           

(4) $\frac{K}{2}\left(l_2^2-l_1^2\right)$