A body of mass m moving with velocity V in the X-direction collides with another body of mass M moving in Y-direction with velocity v. They coalesce into one body during collision. Calculate :
(i) the direction and magnitude of the momentum of the final body.
(ii) the fraction of initial kinetic energy transformed into heat during the collision in terms of the two masses.

Answer

A, B

Explanation

Let V' be the velocity of the final body after collision.
Suppose, V' makes an angle `theta` with x-direction.

(i) Applying conservation of linear momentum in X
direction
`(m+M)V'cos theta=mV….(i)`
Appying conservation of linear momentum in Y
direction
`(m+M)V'sin theta=Mv….(ii)`
Dividing equation (i) and (ii)
`tan theta=(Mv)/(mV) rArr theta=tan^-1((Mv)/(mV))`
This gives the direction of the momentum of the final
body.
Squaring and adding (i) and (ii), we get
`(m+M)^2V'^2cos^2 theta+(m+M)^2V'^2sin^2 theta`
`=m^2V^2+M^2v^2`
`:. V'=sqrt(m^2V^2+M^2v^2)/(m+M)`
Thus the magnitude of the momentum of the final body
`=(m+M)V'=sqrt(m^2V^2+M^2v^2)`
(ii) `(K.E._i-K.E._f)/(K.E._i)=1-(K.E._f)/(K.E._i)=1-(1/2(m+M)V^'2)/[1/2mV^2+M/2v^2]`
`(DeltaKE)/(K.E._i)=1-((m+M)(m^2V^2+M^2v^2)/(m+M^2))/(mV^2+Mv^2)`
`:. (K.E._i-K.E._f)/(K.E._i)=1-(m^2V^2+M^2v^2)/((m+M)(mV^2+Mv^2))`
`=(m^2V^2+mM^2v^2+MmV^2+M^2v^2-mV^2-M^2v^2)/((m+M)(mV^2+Mv^2))`
`(mM(v^2+V^2))/((m+M)(mV^2+Mv^2))`