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Laws of Motion — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsLaws of Motion5 Marks
Question
Two identical billiard balls strike a rigid wall with the same speed but at different angles, and get reflected without any change in speed, as shown in Fig. 4.6. What is (i) the direction of the force on the wall due to each ball? (ii) the ratio of the magnitudes of impulses imparted to the balls by the wall ?
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Answer

An instinctive answer to (i) might be that the force on the wall in case (a) is normal to the wall, while that in case (b) is inclined at 30 to the normal. This answer is wrong. The force on the wall is normal to the wall in both cases.
How to find the force on the wall? The trick is to consider the force (or impulse) on the ball due to the wall using the second law, and then use the third law to answer (i). Let $u$ be the speed of each ball before and after collision with the wall, and $m$ the mass of each ball. Choose the $x$ and $y$ axes as shown in the figure, and consider the change in momentum of the ball in each case :
Case (a)
$
\begin{array}{ll}
\left(p_x\right)_{\text {initial }}=m u & \left(p_y\right)_{\text {iritital }}=0 \\
\left(p_x\right)_{\text {final }}=-m u & \left(p_y\right)_{\text {trial }}=0
\end{array}
$
Impulse is the change in momentum vector. Therefore,
$x$-component of impulse $=-2 mu$
$y$-component of impulse $=0$
Impulse and force are in the same direction. Clearly, from above, the force on the ball due to the wall is normal to the wall, along the negative $x$-direction. Using Newton's third law of motion,
the force on the wall due to the ball is normal to the wall along the positive $x$-direction. The magnitude of force cannot be ascertained since the small time taken for the collision has not been specified in the problem.
Case (b)
$
\left(p_x\right)_{\text {initial }}=m \ u \cos 30^{\circ},\left(p_y\right)_{\text {initial }}=-m \ u \sin 30^{\circ}
$
$
\left(p_x\right)_{\text {final }}=-m \ u \cos 30^{\circ},\left(p_y\right)_{\text {final }}=-m \ u \sin 30^{\circ}
$
Note, while $p_x$ changes sign after collision, $p_y$ does not. Therefore,
$
x \text {-component of impulse }=-2 m \ u \cos 30
$
$
y \text {-component of impulse }=0
$ The direction of impulse (and force) is the same as in (a) and is normal to the wall along the negative $x$ direction. As before, using Newton's third law, the force on the wall due to the ball is normal to the wall along the positive $x$ direction.
The ratio of the magnitudes of the impulses imparted to the balls in (a) and (b) is
$
2 m u /\left(2 m \ u \cos 30^{\circ}\right)=\frac{2}{\sqrt{3}} \approx 1.2
$

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