JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Kinematics

WhatsApp Email

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 slides 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)

Answer: 2

Solution

A to B (45° incline, upward):

By energy conservation, the initial velocity at A:

\frac{1}{2} m v_{0}^{2} = m g h ⟹ v_{0} = 2 g h = 2 \times 10 \times 10 = 102 m/s

Deceleration along the incline:

a = - g sin 45° = - \frac{10}{2}

m/s²

At B, velocity = 0:

0 = v_{0} + a t_{1} ⟹ t_{1} = \frac{10 2}{10/ 2} = 2 s

B to C (30° incline, downward):

Acceleration:

a = g sin 30° = 5

m/s²

Distance along incline:

s = \frac{h}{s i n 30°} = \frac{10}{0.5} = 20

s = \frac{1}{2} a t_{2}^{2} ⟹ 20 = \frac{1}{2} (5) t_{2}^{2} ⟹ t_{2} = 22 s

Total time:

T = t_{1} + t_{2} = 2 + 22 = 2 (1 + 2) s

Comparing with

t (2 + 1)

, we get

t = 2

.

Answer: 2

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Integer answer

Practise more: all JEE Main PYQs on Kinematics →

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Kinematics

Solution

More JEE Main questions on Kinematics