Question

A ball is projected on smooth inclined in direction perpendicular to line of greatest slope with velocity 10 m/s. Choose the correct options(s). (Take g = 10 m/s²)

• A. Angle between initial velocity and velocity at t = 1 s is 45°
• B. Angle between initial velocity and velocity at t = 2 s is 45°
• C. Displacement of ball in 2 s is $10\sqrt{5}$ m
• D. Path of the ball is parabola

Answer 1

Answer: (B), (C), (D)

B, C, D

Answer 2

The ball is projected on a smooth inclined plane with an angle of inclination $\theta = 30°$ and $g = 10$ m/s². The initial velocity is $\vec{v}_0 = 10 \hat{i}$ m/s, perpendicular to the line of greatest slope. The acceleration along the line of greatest slope (y-axis) is $a_y = g \sin\theta = 10 \sin 30° = 5$ m/s². There is no acceleration perpendicular to the line of greatest slope (x-axis), so $a_x = 0$.

The acceleration vector is $\vec{a} = 5 \hat{j}$ m/s². The velocity at time $t$ is $\vec{v}(t) = \vec{v}_0 + \vec{a}t = 10 \hat{i} + 5t \hat{j}$ m/s. The displacement at time $t$ is $\vec{r}(t) = \vec{v}_0 t + \frac{1}{2} \vec{a} t^2 = 10t \hat{i} + \frac{5}{2} t^2 \hat{j}$ m.

A. Angle between initial velocity and velocity at t = 1 s: $\vec{v}(1) = 10 \hat{i} + 5 \hat{j}$. $\cos\phi = \frac{\vec{v}_0 \cdot \vec{v}(1)}{|\vec{v}_0| |\vec{v}(1)|} = \frac{(10 \hat{i}) \cdot (10 \hat{i} + 5 \hat{j})}{10 \sqrt{10^2 + 5^2}} = \frac{100}{10 \sqrt{125}} = \frac{10}{5\sqrt{5}} = \frac{2}{\sqrt{5}}$. This angle is not 45°.

B. Angle between initial velocity and velocity at t = 2 s: $\vec{v}(2) = 10 \hat{i} + 10 \hat{j}$. $\cos\phi = \frac{\vec{v}_0 \cdot \vec{v}(2)}{|\vec{v}_0| |\vec{v}(2)|} = \frac{(10 \hat{i}) \cdot (10 \hat{i} + 10 \hat{j})}{10 \sqrt{10^2 + 10^2}} = \frac{100}{10 \sqrt{200}} = \frac{100}{10 \times 10\sqrt{2}} = \frac{1}{\sqrt{2}}$. So, $\phi = 45°$. This option is correct.

C. Displacement of ball in 2 s: $\vec{r}(2) = 10(2) \hat{i} + \frac{5}{2} (2^2) \hat{j} = 20 \hat{i} + 10 \hat{j}$ m. The magnitude is $|\vec{r}(2)| = \sqrt{20^2 + 10^2} = \sqrt{400 + 100} = \sqrt{500} = 10\sqrt{5}$ m. This option is correct.

D. Path of the ball: $x(t) = 10t \implies t = x/10$. $y(t) = \frac{5}{2} t^2 = \frac{5}{2} (\frac{x}{10})^2 = \frac{5}{2} \frac{x^2}{100} = \frac{1}{40} x^2$. This is the equation of a parabola. This option is correct.