Two particles, A and B, of masses m and 2m, are moving along the X and Y-axis, respectively, with the same speed ϑ.They collide, at the origin, and coalesce into one body, after the collision. What is the velocity of this coalesced mass? What is the loss of energy during this collision
Answers
Answer:
√( mv )^2 + ( 2mv )^2 = 3m × V
√5mv = 3mV
V = √5/3
Answer:
Explanation:
Solution :
Figure shows particles A and B before collision and Figure the coalesced mass after collision.
In figure, we have taken velocity of combined mass (3m) as `vec V` making an angle `alpha` with positive X-axis.
Using law of conservation of linear momentum
(i) along X-axis
`mv=3mVcosalpha` ...(i)
(ii) along Y-axis
`2mv=3mVsinalpha` ...(ii)
Dividing (ii) by (i), we get
`2=tan alpha or alpha = tan^(-1)(2)=63.4^(@)`
Squaring (i) and (ii) and adding, we get
`(3mV)^(2)( cos^(2) alpha + sin^(2)alpha)=m^(2)v^(2)+m^(2)v^(2)`
`=5m^(2)v^(2)`
`V^(2)=(5)/(9)v^(2)or V=( sqrt (5))/(3)v`
No, total K.E. before collision,
`K_(1)=(1)/(2)mv^(2)+(1)/(2)(2m)v^(2)=(3)/(2)mv^(2)`
total K.E. after collision,
`K_(2)=(1)/(2)(3m)V^(2)=(1)/(2)(3m)(5)/(9)v^(2)=(5)/(6)mv^(2)`
Loss of K.E. during the collision,
`Delta K=K_(1)-K_(2)`
` Delta K=(3)/(2)mv^(2)-(5)/(6)mv^(2)=(2)/(3)mv^(2)`
