IIT-JEE 2008 Straight Objective Type (Single Answer Multiple Choice) Questions on Conservation of Energy

The following questions involving the law of conservation of energy are simple even though they may appear to be not so at the first reading:

(1) A block (B) is attached to two unstretched springs S₁ and S₂ with spring constants k and 4 k respectively (see fig.1). The other ends are attached to identical supports M₁ and M₂ not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small distance x (fig. 2) and released. The block returns and moves a maximum distance y towards wall 2. Displacements x and y are measured with respect to the equilibrium position of the block B. The ratio y/x is

(A) 4

(B) 2

(C) ½

(d) ¼

On displacing the block B towards wall 1, spring S₁ gets compressed through x and acquires potential energy ½ kx². When the spring S₁ springs back to its original unstretched condition, it pushes the block B towards wall 2 and compresses the spring S₂ through y. In this compressed condition of spring S₂ the entire kinetic energy of the block B is transferred to spring S₂. Since the spring S₁ is free to move with the block B, it is unstretched and hence we have

½ kx² = ½ (4k) y²

This gives y/x = ½

(2) A bob of mass m is suspended by a massless string of lengh L. The horizontal velocity V at position A is just sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A, satisfies

(A) θ = π/4

(B) π/4 < θ < π/2

(C) π/2 < θ < 3π/4

(D) 3π/4 < θ < π

The sum of the kinetic energy and potential energy of the of the bob in the displaced position must be equal to the kinetic energy at the position A. Therefore we have

½ M (V/2)² + MgL(1 – cos θ) = ½ MV²

The critical velocity (for just tracing the vertical circle) V = √(5gR) = √(5gL)

Substituting this value we obtain

gL(1 – cos θ) = 15gL/8 so that cos θ = – 7/8

Therefore 3π/4 < θ < π.