Question

A small ball is projected up a smooth inclined plane with an initial speed of 10 m/s along the direction at 30° to the bottom edge of the slope. It returns to the edge after 2 s. The ball is in contact with the inclined plane throughout the process. The angle of inclination is given by $\frac{\pi}{\beta}$. Then $\beta$ is

Answer 1

6

Answer 2

Let the angle of inclination of the smooth inclined plane with the horizontal be $\theta$. Let the initial speed of the ball be $u = 10$ m/s. The initial velocity is along the direction at an angle $\alpha = 30^\circ$ to the bottom edge of the slope. We can set up a coordinate system on the inclined plane. Let the x-axis be along the bottom edge of the slope and the y-axis be perpendicular to the bottom edge, lying on the inclined plane. The initial velocity vector lies on the inclined plane. The components of the initial velocity are $u_x = u \cos \alpha$ along the x-axis and $u_y = u \sin \alpha$ along the y-axis. $u_x = 10 \cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ m/s. $u_y = 10 \sin 30^\circ = 10 \times \frac{1}{2} = 5$ m/s.

The only force acting on the ball parallel to the inclined plane is the component of gravity along the inclined plane, which is $g \sin \theta$, directed downwards along the steepest descent line. The steepest descent line is perpendicular to the bottom edge on the inclined plane, which is along the negative y-direction in our chosen coordinate system. The acceleration components on the inclined plane are $a_x = 0$ (since there is no force component along the bottom edge) and $a_y = -g \sin \theta$.

The ball is projected from the edge (y=0) and returns to the edge (y=0) after $t = 2$ seconds. We can use the equation of motion in the y-direction: $y = u_y t + \frac{1}{2} a_y t^2$. When the ball returns to the edge, $y = 0$ at $t = 2$ s. $0 = (5)(2) + \frac{1}{2} (-g \sin \theta) (2)^2$ $0 = 10 - \frac{1}{2} g \sin \theta \times 4$ $0 = 10 - 2 g \sin \theta$ $2 g \sin \theta = 10$ $g \sin \theta = 5$.

In typical physics problems at this level, when $g$ is not explicitly given, it is often taken as 10 m/s$^2$. Assuming $g = 10$ m/s$^2$: $10 \sin \theta = 5$ $\sin \theta = \frac{5}{10} = \frac{1}{2}$. Since $\theta$ is the angle of inclination of the plane, it is positive and usually less than or equal to $\pi/2$. The principal value of $\theta$ such that $\sin \theta = \frac{1}{2}$ is $\theta = 30^\circ = \frac{\pi}{6}$ radians.

We are given that the angle of inclination is $\theta = \frac{\pi}{\beta}$. Comparing this with our result, $\frac{\pi}{\beta} = \frac{\pi}{6}$. Therefore, $\beta = 6$.

Let's verify the assumption $g=10$ m/s$^2$. If we did not assume $g=10$, we would have $\sin \theta = \frac{5}{g}$. Then $\frac{\pi}{\beta} = \arcsin \left( \frac{5}{g} \right)$, so $\beta = \frac{\pi}{\arcsin \left( \frac{5}{g} \right)}$. Since the problem asks for a specific value of $\beta$, and the value 6 obtained with $g=10$ leads to a simple angle $\theta = 30^\circ$, it is highly probable that $g=10$ was intended. Also, the time of flight calculation confirms this: time to reach maximum y-displacement is $t_{top} = \frac{-u_y}{a_y} = \frac{-5}{-g \sin \theta} = \frac{5}{g \sin \theta} = \frac{5}{5} = 1$ s. The total time of flight is $2 \times t_{top} = 2 \times 1 = 2$ s, which matches the given time.