2019 Laws of MotionMotion on an Inclined Plane Medium Question 2 of 2 in Motion on an Inclined Plane

When an object is shot from the bottom of a long smooth inclined plane kept at an angle $60^\circ$ with horizontal, it can travel a distance $x_1$ along the plane. But when the inclination is decreased to $30^\circ$ and the same object the shot with the same velocity, it can travel $x_2$ distance. Then $x_1:x_2$ will be:

A $1:\sqrt{2}$
B $\sqrt{2}:1$
C $1:\sqrt{3}$
D $1:2\sqrt{3}$

The answer is C.

Solution

On a smooth incline the retardation is $g\sin\theta$, so the stopping distance is $x=\frac{u^2}{2g\sin\theta}$. $x_1=\frac{u^2}{2g\sin60^\circ}$, $x_2=\frac{u^2}{2g\sin30^\circ}$ $\frac{x_1}{x_2}=\frac{\sin30^\circ}{\sin60^\circ}=\frac{1/2}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}$ $x_1:x_2=1:\sqrt{3}$

Concept distance travelled up an incline at different angles

Motion on an Inclined Plane in past papers

2 questions from this chapter have appeared across 2 exam years.