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?

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:

Let the mass of each ball be m kg

v be the velocity of ball A along positive x axis

v 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

∴  v = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v

2 = u +  v → equation 1

∴ relative velocity of approach = relative velocity of separation

-2 =  v - u → equation 2

By adding both equations 1 and 2 we get

v = 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

hope it

i think the correct answer from the choices listed above is option c. when the kinetic energy of particles in a liquid state decreases

explanation: