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Question: Two identical billiard balls strike a rigid wall with the same speed but at different angles, and ge...

Two identical billiard balls strike a rigid wall with the same speed but at different angles, and get reflected without any loss of speed as shown in the figure. What is (i)\left( i \right) the direction of the force of the wall due to each ball? And (ii)\left( ii \right) the ratio of magnitudes of the impulses imparted on the two balls by the wall?

Explanation

Solution

To solve the parts of the given question, the concept of impulse must be thorough. Impulse of a force is the total effect of a large force which acts for a short time to produce a finite change in momentum and is physically the integral of a force over the time interval.

Complete step by step answer:
Primarily, to find the direction of the force of the wall due to each ball,
Let uu be the speed of each ball before and after collision with the wall, and mm be the mass of each ball and then choose the yaxisy-axis and xaxisx-axis as given in the above picture.
In according to the question,
In x component,
Initial momentum,pxinitial=mup_{x}^{initial}=mu and final momentum pxfinal=mup_{x}^{final}=-mu
In y component,
Initial momentum,pyinitial=0p_{y}^{initial}=0 and final momentum pyfinal=0p_{y}^{final}=0
As impulse=change in momentumimpulse=change\text{ }in\text{ }momentum
Therefore, xcomponentof the impulse=mumu=2mux-component of\text{ }the\text{ }impulse=-mu-mu=-2mu
And ycomponentsof the impulse=0y-components of\text{ }the\text{ }impulse=0
It is clear that the direction of the impulse is along the negative xx-direction of the motion and as the direction of the force is the same as that of impulse, so the force on the ball due to the wall is normal to the wall and along the xx-direction of the motion. Thus by Newton’s third law of motion, the force on the wall due to the ball is normal to the wall along the positive xx-direction of the motion.
Now, for the second ball incident at the wall at 30{{30}^{\circ }}
In x component,
Initial momentum,pxinitial=mucos30p_{x}^{initial}=mu\cos {{30}^{\circ }} and final momentum pxfinal=mucos30p_{x}^{final}=-mu\cos {{30}^{\circ }}
In y component,
Initial momentum,pyinitial=musin30p_{y}^{initial}=-mu\sin {{30}^{\circ }} and final momentum pyfinal=musin30p_{y}^{final}=-mu\sin {{30}^{\circ }}
Therefore, xcomponent of the impulse=mucos30mucos30=2mucos30x-component\ of\text{ }the\text{ }impulse=-mu\cos {{30}^{\circ }}-mu\cos {{30}^{\circ }}=-2mu\cos {{30}^{\circ }}
And ycomponents of the impulse=musin30+musin30=0y-components\ of\text{ }the\text{ }impulse=-mu\sin {{30}^{\circ }}+mu\sin {{30}^{\circ }}=0
Again, the direction of the impulse is normal to the wall along the negative xx-direction and by Newton’s third law of motion, the force on the wall due to the ball is normal to the wall along the positive xx-direction.
Secondarily, we need to find the ratio of magnitudes of the impulses imparted on the two balls by the wall; we need to find the ratio of magnitudes of xcomponentof x-component of\text{ } the impulse imparted on the two balls by the wall i.e.
pa:pb=2mu:mucos30{{p}_{a}}:{{p}_{b}}=-2mu:-mu\cos {{30}^{\circ }}
pa:pb=2:32\Rightarrow {{p}_{a}}:{{p}_{b}}=-2:\frac{\sqrt{3}}{2}
pa:pb=4:3\Rightarrow {{p}_{a}}:{{p}_{b}}=4:\sqrt{3}

Therefore the ratio of the impulses imparted on the two balls by the wall is 4:34:\sqrt{3}.

Note:
One is advised to carefully observe that only the ycomponent of y-component\ of\text{ } impulse imparted on the two balls due to the wall while calculating the ratio of magnitudes of the impulses imparted on the two balls by the wall as the ycomponentof y-component of\text{ } the impulse is null in both the cases.