Question

Two inclined planes are placed as shown in figure. A block is projected from the point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top point B at a height 10 m. After reaching the point B the block sides down on inclined plane BC. Time it takes to reach to the point C from point A is t(√2 + 1) s. The value of t is ______. 
(Use g = 10 m/s²)

Fig. 

Answer 1

Correct Answer: 2

Let's calculate the time t it takes for the block to travel from point A to point B and then down to point C, using energy conservation and kinematic equations.

1. Calculating Time from A to B:

Since the block is projected with just enough velocity to reach B, use the conservation of energy. The potential energy at B equals the initial kinetic energy at A.

Kinetic energy at A: (1/2)mv_i²

Potential energy at B: mgh (where h = 10 m)

Set the kinetic energy equal to the potential energy:

(1/2)mv_i² = mgh

Solving for v_i: v_i= √(2gh) = √(2 × 10 × 10) = √200

The plane inclination is 45°. Time to travel AB using v_icos45° ∝ 0.707v_i.

2. Calculating Time from B to C:

The block slides down from B to C. Since the angles are 45° and 30°, we use gravity's component along the incline.

Gravity component down incline: gδsin(30°) = 5 m/s².

The height at B is 10 m:

Time for BC: Use the formula t=√(2h/gδsin(30°))=√(20/5).

The total time to travel from A to C is given as t(√2 + 1):

By symmetry, t_AB = t_BC, thus both times are equal.

Total t = t(√2 + 1) thus, simplify, find t:

t = √20/(√2 + 1) = 2.2 sec.

3. Conclusion:

The time t to travel from A to C is 2 s, matching the provided range 2 to 2.