Question

A particle strikes a horizontal frictionless floor with a speed $u$, at an angle $\theta$ to the vertical, and rebounds with a speed $v$, at an angle $\phi$ to the vertical. The coefficient of restitution between the particle and the floor is $e$. The magnitude of $v$ is

In the previous question the angle $\phi$ is equal to

• A. $eu$
• B. $(1-e)u$
• C. $u\sqrt{sin^2\theta + e^2cos^2\theta}$
• D. $u\sqrt{e^2sin^2\theta + cos^2\theta}$

Answer 1

Answer: (C)

c

Answer 2

The problem involves an oblique collision of a particle with a horizontal frictionless floor. We need to determine the magnitude of the rebound velocity ($v$) and the angle of rebound ($\phi$) with the vertical.

Let the initial velocity be $\vec{u}$ and the final velocity be $\vec{v}$. The angles $\theta$ and $\phi$ are given with respect to the vertical.

1. Resolve velocities into components: Let the horizontal direction be x and the vertical direction be y. Initial velocity components:

• Horizontal component: $u_x = u \sin\theta$
• Vertical component: $u_y = u \cos\theta$ (downwards)

Final velocity components:

• Horizontal component: $v_x = v \sin\phi$
• Vertical component: $v_y = v \cos\phi$ (upwards)

2. Apply collision principles:

• Conservation of momentum in the horizontal direction: Since the floor is frictionless, there is no impulsive force in the horizontal direction. Therefore, the horizontal component of the particle's velocity remains unchanged. $u_x = v_x$ $u \sin\theta = v \sin\phi \quad \dots (1)$

• Coefficient of restitution ($e$) in the vertical direction: The coefficient of restitution relates the relative velocity of separation to the relative velocity of approach along the common normal (which is the vertical direction here). The floor is stationary. $e = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}} = \frac{v_y}{u_y}$ $v_y = e u_y$ $v \cos\phi = e u \cos\theta \quad \dots (2)$

3. Calculate the magnitude of $v$: To find $v$, square equations (1) and (2) and add them: $(v \sin\phi)^2 + (v \cos\phi)^2 = (u \sin\theta)^2 + (e u \cos\theta)^2$ $v^2 \sin^2\phi + v^2 \cos^2\phi = u^2 \sin^2\theta + e^2 u^2 \cos^2\theta$ $v^2 (\sin^2\phi + \cos^2\phi) = u^2 (\sin^2\theta + e^2 \cos^2\theta)$ Using the trigonometric identity $\sin^2\alpha + \cos^2\alpha = 1$: $v^2 = u^2 (\sin^2\theta + e^2 \cos^2\theta)$ $v = u \sqrt{\sin^2\theta + e^2 \cos^2\theta}$

4. Calculate the angle $\phi$: To find $\phi$, divide equation (1) by equation (2): $\frac{v \sin\phi}{v \cos\phi} = \frac{u \sin\theta}{e u \cos\theta}$ $\tan\phi = \frac{1}{e} \frac{\sin\theta}{\cos\theta}$ $\tan\phi = \frac{1}{e} \tan\theta$ $\phi = \tan^{-1}\left[\frac{1}{e} \tan\theta\right]$

The magnitude of $v$ is $u\sqrt{\sin^2\theta + e^2\cos^2\theta}$, which corresponds to option (c) for the first part. The angle $\phi$ is $\tan^{-1}\left[\frac{1}{e}\tan\theta\right]$, which corresponds to option (c) for the second part.