Question

A particle of mass m with an initial velocity u`hat"i"` collides perfectly elastically with a mass 3m at rest. It moves with a velocity v`hat"j"` after collision, then, v is given by :

पर्याय

• v = `sqrt(2/3"u")`

• v = `"u"/sqrt3`

• v = `1/sqrt6"u"`

• v = `"u"/sqrt2`

Answer 1

`bb("v" = "u"/sqrt2)`

Explanation:

Given: The mass of particle 1 is m, and its initial velocity is u`hat"i"`. The mass of particle 2 is 3m, and its initial velocity is 0; the velocity of particle 1 after collision is v`hat"j"`.

To find: Magnitude of v.

Let velocity of particle - 2 after collision be v'

By law of conservation of linear momentum:

mu`hat"i"` + 0 = mv `hat"j"`+ 3mv'

v' = `"u"/3hat"i" -"v"/3hat"j"`

By law of conservation of kinetic energy:

`1/2"mu"^2+0=1/2"mv"^2+1/2(3"m")"v'"`

u² = `"v"^2+3[("u"/3)^2+("v"/3)^2]`

`"u"^2-"u"^2/3 = "v"^2+"v"^2/3`

`2/3"u"^2=4/3"v"^2`

u² = 2v²

v = `"u"/sqrt2`