Question

When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60 degree with horizontal ,it can travel a distance x1 along the plane .But when the inclination is decreased to 30 degree and the same object is shot with the same velocity ,it can travel x2 distance . Then x1:x2 will be a)root2:1 b)1:root3 c)1:2root3 d)1:root2

Answer 1

The ratio of x₁ to x₂ is 1/√3

Given-

• Angle of inclined plane = 60°
• Along the plane distance travelled is = x₁
• Inclination decreased upto = 30°
• When inclination is 30° then distance travelled is = x₂

Here initial velocity (u) is same for both the cases.

From the attached figure it is clear that acceleration can be given by

a = -g sinθ

From Newtons laws of motion

v² = u² + 2as

where S is the distance and a is the acceleration.

For first case

0 = u² + 2 (-g sinθ₁)x₁

x₁ = u²/2g sinθ₁

For second case

0 = u² + 2 (-g sinθ₂)x₂

x₂ = u²/2g sinθ₂

By substituting the values we get

x₁ = u²/2g sin60°

x₂ = u²/2g sin30°

x₁/x₂ = sin 30°/sin 60° = 1/√3

Answer 2

Question: When an object is shot from the bottom of a long smooth inclined plane kept at an angle $60^{o}$ with horizontal, it can travel a distance $x1$ along the plane. But when the inclination is decreased to $30^{o}$ and the same object is shot with the same velocity, it can travel $x2$ distance. Then $x1:x2$ will be:

Solution: $1:\sqrt{3}$

Step by step explanation: (Stopping distance) $x_1=\frac{u^{2} }{2gsin60^{o} }$

(Stopping distance) $x_2=\frac{u^{2} }{2gsin30^{o} }$

$:\implies\frac{x_1}{x_2} =\frac{sin30^{o} }{sin60^{o} }=\frac{1 \times 2}{2 \times \sqrt{3} } = 1:\sqrt{3}$

Hence, $x1:x2 = \boxed{1:\sqrt{3} }$