Question

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is moving upward along the y axis at 2.0 m/s and ball B is moving to the right along the x axis with speed 3.7 m/s. After the collision ball B is moving along the positive y axis. a. What is the speed of ball A and B after the collision? b. What is the direction of motion of ball A after the collision? b. What is the total momentum and kinetic energy of the two balls after the collision?

Answer 1

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system

Let the mass of each ball be m kg

v$_{1}$ be the velocity of ball A along positive x axis

v$_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v$_{1}$

∴  v$_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v$_{2}$

2 = u +  v$_{2}$ → equation 1

Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision

And for an elastic collision, coefficient of restitution = 1

∴ relative velocity of approach = relative velocity of separation

-2 =  v$_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v$_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

                         = 8·85×m J

Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s