A uniform rod of mass m and length 2a lies at rest on a smooth horizontal table. A perfectly elastic particle of same mass m, moving with speed `upsilon` on the table in a direction perpendicular to the rod, strikes one end of the rod. The kinetic energy generated in the rod is given by `n/25 m upsilon^2`.Find the value of n.

Answer

8

Explanation

From conservation of linear momentum we have
`v = v_1 + v_2 " " .....(1)`
From conservation of angular momentum about centre of rod we have,
`mva = mv_2 a + (ma^2)/3 omega " or " v = v_2 + (a omega)/3 " " ...(2)`
Further from the definition of coefficient of restitution(e=1) at point of impact.
Relative speed of approach=relative speed of separation.
`therefore v = v_1 + a omega - v_2 " " .....(3) `
Solving these eqns. (1), (2) and (3) we get,
`v_1 = 2/5 v and omega = (6v)/(5a)`
`therefore` Kinetic energy of rod, `K = 1/2 xx m xx (2/5 v)^2 + 1/2 xx (ma^2)/3 xx ((6 v)/(5a))^2 = 8/25 mv^2 therefore n = 8`