A 0.20-kg billiard ball traveling at a speed of 15 m/s strikes the side rail of a pool table at an angle of 60o. If the ball rebounds at the same speed and angle, what is the change in its momentum?
The velocity component at right angles to the side rail reverses sign in theOriginally Posted by Mohycha
collision. If we take positive upwards and to the right the initial velocity of
the ball is:
$\displaystyle 15\ \left( \frac{1}{2\sqrt{3}},\frac{1}{2}\right)$,
and the final velocity is:
$\displaystyle 15\ \left( \frac{1}{2\sqrt{3}},-\frac{1}{2}\right)$.
Therefore the change in velocity is:
$\displaystyle 15\ (0,-1)$,
and (as the mass of the ball does not change) the change in momentum is:
$\displaystyle 0.2\times 15\ (0,-1)=(0,-3)$.
RonL