Two particles, A and B of masses m and 3m, are moving along X and Y axes respectively, with the same speed v. They collide at the origin, and coalesce into one body, after the c ollision. What is the velocity of this coalesced mass ?

Explanation

Hint :
Let V be the speed of the combined mass and let V make an angle `alpha` with `+ve` x - axis
`mv=4mv cos alpha" …(i) (conservation of momentum for x component)"`
`3mv=4mV sin alpha" … (ii)"`
Find `tan alpha`,
Find `cos alpha and sin alpha` from (i) & (ii) and put in `cos^(2)alpha +sin^(2)alpha=1`
This will give V in terms of v

Figure shows the two particles before and after the collision. Let V be the speed of the combined mass and let the direction of `vecV` be making an angle `alpha` with the positive X - axis, after collision, Use law of conservation of linear momentum
`mv=4mV cos alpha" ...(i) (For X - components)"`
`3mv=4mV sin alpha" ...(ii) (For Y - components)"`
Divide equation (ii) by equation (i),
`(4mV sinalpha)/(4mV cos alpha)=(3mv)/(mv)`
`tan alpha=3" ...(iii)"`
From (i), `cos alpha=(v)/(4V)" ...(iv)"`
`"From (ii), "sin alpha=(3v)/(4V)" ...(v)"`
`cos^(2)alpha+sin^(2)alpha=1" ...(vi)"`
Put values from (iv) & (v) in (vi)
`((v)/(4V))^(2)+((3v)/(4V))^(2)=1`
`V^(2)=((v)/(4))^(2)+((3v)/(4))^(2)=(10)/(16)v^(2)=(5)/(8)v^(2)`
`V=sqrt((5)/(8))v`