Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. One is moving upward along the y-axis at 3.00 m/s, the other is moving to the right along the x-axis with speed 4.80 m/s. After the collision (assumed elastic), the second ball is moving along the positive y-axis. What is the final direction of the first ball and what are their two speeds after the collision.
Can someone please show me step-by-step how to solve this problem? Thank you!
1 Answer
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let mass = 1 kg for simplicity
momentum of first is 3 kgm/s in y direction
momentum of second is 4.8 kgm/s in x direction.
After collision the total momentum has to match that before the collision in both x and y components.
total momentum is √(3² + 4.8²) = 5.66
Assume angle a for #1, angle with the +x axis
then total y component is V1sina + V2
and total x component is V1cosa
set them equal to initial x and y components
V1sina + V2 = 3
V1cosa = 4.8
total momentum is
√((V1sina+V2)² + (V1cosa)²) = 5.66
(V1sina+V2)² + (V1cosa)² = 32.04
V1²sin²a + V2² + 2V1V2sina + V1²cos²a = 32.04
V1²(sin²a+cos²a) + V2² + 2V1V2sina = 32.04
V1² + V2² + 2V1V2sina = 32.04
V1sina = 3 – V2
V1² + V2² + 2V2(3 – V2) = 32.04
V1² + V2² + 6V2 – 2V2² = 32.04
V1² – V2² + 6V2 = 32.04
3 equations 3 unknowns
V1² – V2² + 6V2 = 32.04
V1sina + V2 = 3
V1cosa = 4.8
intuitively I know that a is zero, let’s see if that is a solution.
plugging that in
V2 = 3
V1 = 4.8
does that solve the 3rd equation?
V1² – V2² + 6V2 = 32.04
23.04 – 9 + 18 = 32.04
yes.
