Question

A billiard ball with a speed of 5 m/s strikes another stationary billiard ball in a perfectly elastic collision.

After the collision, the first ball has a speed of 4.35 m/s and is traveling at an angle of 30° below its original line of motion.

Find the speed and angle of the second ball, with respect to the initial line of motion, after the collision.

Answer 1

Assuming both billiard balls have the same mass, conservation of momentum says

$m\vec v_1 + m\vec v_2 = m{\vec v_1}\,' + m{\vec v_2}\,'$

where m = mass of both billiard balls, and v₁ and v₂ = their initial velocities, and v₁' and v₂' = their final velocities. The masses are the same so the exact value of m is irrelevant. The first ball has initial speed 5 m/s and the second is at rest, so

$\left(5 \dfrac{\rm m}{\rm s}\right) \, \vec\imath = {\vec v_1}\,' + {\vec v_2}\,'$

After the collision, the first ball has speed 4.35 m/s and is moving at angle of 30° below the original path, so

${\vec v_1}\,' = \left(4.35\dfrac{\rm m}{\rm s}\right)\left(\cos(30^\circ) \, \vec\imath + \sin(30^\circ) \, \vec\jmath\right) \approx \left(3.77 \dfrac{\rm m}{\rm s}\right) \vec\imath + \left(-2.18 \dfrac{\rm m}{\rm s}\right) \vec\jmath$

Then the second ball has final velocity vector

${\vec v_2}\,' \approx \left(1.23 \dfrac{\rm m}{\rm s}\right) \vec\imath + \left(2.18 \dfrac{\rm m}{\rm s}\right) \vec\jmath$

so it moves with speed

$\left\|{\vec v_2}\,'\right\| \approx \sqrt{\left(1.23\dfrac{\rm m}{\rm s}\right)^2 + \left(2.18\dfrac{\rm m}{\rm s}\right)^2} \approx \boxed{2.50 \dfrac{\rm m}{\rm s}}$

at an angle of

$\theta \approx \tan^{-1}\left(\dfrac{2.18}{1.23}\right) \approx \boxed{60.5^\circ}$

or about 60.5° above the original line of motion.