Respuesta :
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[tex]_{1}[/tex] be the velocity of ball A along positive x axis
v[tex]_{2}[/tex] 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[tex]_{1}[/tex]
∴ v[tex]_{1}[/tex] = 3.7 m/s along positive x axis
Conservation of momentum along y axis
m×2 = m×u + m× v[tex]_{2}[/tex]
2 = u + v[tex]_{2}[/tex] → 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[tex]_{2}[/tex] - u → equation 2
By adding both equations 1 and 2 we get
v[tex]_{2}[/tex] = 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
