Question

A small spherical ball undergoes an elastic collision with a rough horizontal surface. Before the collision, it is moving at an angle $θ$ to the horizontal (see figure). You may assume that the frictional force obeys the law $f = μ N$ during the contact period, where $N$ is the normal reaction on the ball and $μ$ is the coefficient of friction.

$(a)$ Obtain $θ_{m} (μ)$ , so that the subsequent horizontal range of the ball after leaving the horizontal surface is maximised.

$(b)$ Find the allowed range for $θ_{m} .$

Accepted Answer

Refer the diagram above, the velocity is resolved in terms of horizontal and vertical components as shown.
Now we have to consider the range which is given by,
$R = v_{x} \times T$ , where $v_{x}$ is the horizontal component of the velocity and $T$ is the time of flight.
Now applying the concept of Newton's second law,
Along vertical axis,
$∆ p = F t 2 m v_{y} \sin (θ) = N t$
where $N$ is the normal reaction.
Along horizontal axis,
$m v \cos (θ) - μ N t = m v_{x}$
Putting value of $N t$ in the above equation,
$m v \cos (θ) - μ \times (2 m v_{y} \sin (θ)) = m v_{x} \Rightarrow v_{x} = v \cos (θ) - 2 v μ \sin (θ)$

Now the range is given by,
$R = [v \cos (θ) - 2 v μ \sin (θ)] \times (\frac{2 v \sin (θ)}{g}) \Rightarrow R = \frac{2 v^{2}}{g} \times \sin (θ) (\cos (θ) - 2 μ \sin (θ))$
For maximum range differentiating above equation we get,
$\frac{d R}{d θ} = 0 \Rightarrow \frac{d}{d θ} [\sin (θ) (\cos (θ) - 2 μ \sin (θ))] = 0 \Rightarrow \frac{d}{d θ} [\frac{\sin (2 θ)}{2} - 2 μ (\frac{1 - \cos (2 θ)}{2})] = 0 \Rightarrow \cos (2 θ) = 2 μ \sin (2 θ) \Rightarrow \tan (2 θ) = \frac{1}{2 μ} \Rightarrow θ_{m} = \frac{1}{2} \cot^{- 1} (2 μ)$
The allowed range for the value of $θ_{m}$ should be,
$0 \leq θ_{m} \leq \frac{π}{4}$

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