A particle free to move along the x-axis has potential energy given by $U\left(x\right)=k[1-e^{-x^2}]$ for $-\infty \le x\le +\infty$, where k is a positive constant of appropriate dimensions. Then 

(1) At point away from the origin, the particle is in unstable equilibrium

(2) For any finite non-zero value of x, there is a force directed away from the origin

(3) If its total mechanical energy is k/2, it has its minimum kinetic energy at the origin

(4) For small displacements from x = 0, the motion is simple harmonic

The potential energy of a system is represented in the first figure. the force acting on the system will be represented by

(1)

(2)

(3)

(4)

The force acting on a body moving along x-axis varies with the position of the particle as shown in the fig.

The body is in stable equilibrium at

(1) x = x₁

(2) x = x₂

(3) both x₁ and x₂

(4) neither x₁ nor x₂

The potential energy of a particle varies with distance x as shown in the graph. The force acting on the particle is zero at

(1) C

(2) B

(3) B and C

(4) A and D

A particle is placed at the origin and a force F = kx is acting on it (where k is positive constant). If U(0) = 0, the graph of U(x) versus x will be (where U is the potential energy function) 

(1)

(2)

(3)

(4)

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