A ball collides elastically with another ball of the same mass. The collision is oblique and initially one of the balls was at rest. After the collision, the two balls move with same speeds. What will be the angle between the velocity of the ball after the collision?

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. `90^@`

Answer

B

Explanation

Applying law of conservation of momentum along horizontal and vertical directions, we get,
`m v sin theta_1 - m v sin theta_2 = 0`
i.e., `theta_1 = theta_2 " " ...(1)`
Also , `m u = m v ( cos theta_1 + cos theta_2) = 2 mv cos theta [theta_1 = theta_2 = theta"(say)"] `
`cos theta = u/(2v) " " .....(2)`
According to law conservation of kinetic KE, `1/2 m u^2 = 1/2 mv^2 + 1/2 mv^2`
`" or " u^2 = 2v^2 " or " u = sqrt(2) v " " ....(3)`
From eqn. (2) and eqn. (3), `cos theta = (sqrt(2)v)/(2v) " or ", cos theta = 1/(sqrt(2)) " or " theta = 45^@`
Hence `theta_1 = theta_2 = 45^@`