What is the final direction of ball A? What are their two speeds?

Question

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Ball A is moving upward along the y axis at vA = 2.4 m/s , and ball B is moving to the right along the x axis with speed vB = 5.2 m/s . After the collision, assumed elastic, ball B is moving along the positive y axis.

Answer 1

In an elastic collision, I believe the balls would exchange velocities.

If a ball of velocity v strikes a ball of equal mass but velocity of zero, the first ball would stop and the second would continue with the velocity of the first in an elastic collision. The elastic condition, (delta)v-initial ,=-(delta)v-final, where the (delta) is between A and B, demands it.

Now picture the horizontal interaction: between the two balls. B has a horizontal speed of 5.2, B has a horizontal speed of zero. They then exchange horizontal velocities: now A has a horizontal speed of 0 and B has a horizontal speed of 5.2

Same for the vertical velocities: the new vertical velocity of A is 0 and the new vB is 2.4

Last bit of proof: KE of the system before collision is the sum of the KE of each ball, or

0.5(m)(5.2 m/s)^2 + 0.5(m)(2.4 m/s)^2.

After the collision, it's

0.5(m)(2.4 m/s)^2 + 0.5(m)(5.2 m/s)^2

Hey, those are equal! KE is conserved, as is demanded in an elastic collision.

Answer 2

The problem allows us to assume that the collision is an elastic collision. An elastic collision is a collision where both kinetic energy, KE, and momentum, p, are conserved. Therefore KE_(initial) = KE_(final) and p_(initial) = p_(final) , where the "initial" subscript denotes the value before the collision and the "final" subscript denotes the vale after the collision. Kinetic Energy is related to mass and velocity by this equation: KE = 1/2 mv² . Linear momentum, p, is related to mass and velocity by p = mv. Using these relationships, we can derive the equations we need to solve the problem:

[1] p_{(initial, ball A)} + p_{(initial, ball B)} = p_{(final, ball A)} + p_{(final, ball B)}

[2] KE_{(initial, ball A)} + KE_{(initial, ball B)} = KE_{(final, ball A)} + KE_{(final, ball B)}

Next we substitute the equations for p and KE, such that equation [1] and [2] are in terms of mass (m) and velocity (v):

[1] m_{(ball A)}v_{(initial, ball A)} + m_{(ball B)}v_{(initial, ball B)} = m_{(ball A)}v_{(final, ball A)} + m_{(ball B)}v_{(final, ball B)}

[2] ½ m_{(ball A)}v²_{(initial, ball A)} + ½ m_{(ball B)}v²_{(initial, ball B)} = ½ m_{(ball A)}v²_{(final, ball A)} + ½ m_{(ball B)}v²_{(final, ball B)}

The problem gives us "vA = 2.4 m/s" and "vB = 5.2 m/s", "two billiard balls of equal mass", and "Ball A is moving upward along the y axis...Ball B is moving to the right along the x axis". We'll use the x axis as our reference direction. We can rewrite these expressions using the nomenclature adopted in the above equations for consistency:

vA = v_{(initial, ball A)} = 2.4 m/s ∠90°

vB = v_{(initial, ball B)} = 5.2 m/s

m_{(ball A)} = m_{(ball B)} = m (since these are equal, the subscript for mass values is no longer needed)

Now we'll input these relationships into equations [1] and [2]:

[1] m*2.4 m/s ∠90° + m*5.2 m/s = mv_{(final, ball A)} + mv_{(final, ball B)}

[2] ½ m*(2.4 m/s ∠90°)² + ½ m*(5.2 m/s)² = ½ mv²_{(final, ball A)} + ½ mv²_{(final, ball B)}

From here we can use our algebra knowledge to clearly see that m (mass) can be factored out and cancelled from equations [1] and [2] and "1/2" can be factored out and cancelled from equation [2]. This simplification leaves us with:

[1] 2.4 m/s ∠90° + 5.2 m/s = v_{(final, ball A)} + v_{(final, ball B)}

[2] (2.4 m/s ∠90°)² + (5.2 m/s)² = v²_{(final, ball A)} + v²_{(final, ball B)}

We now have two equation with two unknowns. Note: Since this problem has a directional (or vector/phasor) component, it's advised to sketch these graphically on the xy coordinate system proposed in the problem. This way, the answer to the final direction of ball A can be seen graphically, in addition to being represented with an angle value relative to the x-axis.

Note that the problem statement told us "After the collision...ball B is moving along the positive y axis", and our reference is the x-axis. So, equations [1] and [2] can be visually solved for v_{(final, ball A)} and v_{(final, ball B)} with this knowledge:

v_{(final, ball A)} = 5.2 m/s

v_{(final, ball B)} = 2.4 m/s ∠90°

Per our answers, the final direction of ball A is along the positive x axis and the two speeds are 5.2 m/s for ball A and 2.4 m/s for ball B.

Hope this helps and Good Luck!

-Michael Mount