Question 8.3: Two identical billiard balls strike a rigid wall with the sa...

Solution for (a)
The first ball bounces directly into the wall and exerts a force on it in the +x direction. Therefore the wall exerts a force on the ball in the −x direction. The second ball continues with the same momentum  component in the y direction, but reverses its x -component of  momentum, as seen by sketching a diagram of the angles involved and keeping in mind the proportionality between velocity and momentum. These changes mean the change in momentum for both balls is in the −x direction, so the force of the wall on each ball is along the −x direction.

Solution for (b)
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 -axis and y -axis as previously described, and consider the change in momentum of the first ball which strikes perpendicular to the wall.

p_{xi}  =  mu;  p_{yi}  =  0                   (8.19)

p_{xf}  =  −mu;  p_{yf}  =  0                    (8.20)
Impulse is the change in momentum vector. Therefore the x  -component of impulse is equal to −2mu and the y -component of impulse is equal to zero. Now consider the change in momentum of the second ball.

p_{xi}  =  mu  cos 30º;  p_{yi} = –mu  sin 30º                  (8.21)

p_{xf}  =  – mu  cos 30º;  p_{yf}  =  −mu  sin 30º                     (8.22)

It should be noted here that while p_{x} changes sign after the collision, p_{y} does not. Therefore the x -component of impulse is equal to −2mu cos 30º and the y -component of impulse is equal to zero. The ratio of the magnitudes of the impulse imparted to the balls is

\frac{2mu}{2mu cos 30º} = \frac{2}{\sqrt{3} } = 1.155.                   (8.23)

Discussion
The direction of impulse and force is the same as in the case of (a); it is normal to the wall and along the negative x -direction. Making use of Newton’s third law, the force on the wall due to each ball is normal to the wall along the positive x -direction.