JEE Main 2014 · 12 April · Online · Physics · Center of Mass, Momentum and Collision
Three masses m, 2 ~m and 3 ~m are moving in x-y plane with speed 3...
Three masses m,2m and 3m are moving in x−y plane with speed 3u,2u and u respectively as shown in figure. The three masses collide at the same point at P and stick together. The velocity of resulting mass will be:
Options
A12u(i^+3j^)
B12u(i^−3j^)
C12u(−i^+3j^)
D12u(−i^−3j^)
Correct answer
D. 12u(−i^−3j^)
Step-by-step solution
From the law of conservation of momentum we know that,
\mathrm{m}}_{1} \mathrm{u}}_{1}+\mathrm{m}}_{2} \mathrm{u}}_{2}+\ldots .=\mathrm{m}}_{1} \mathrm{v}}_{1}+\mathrm{m}}_{2} \mathrm{v}}_{2}+\ldots .
Given m1=m,m2=2m and m3=3m and u1=3u,u2=2u and u3=u
Let the velocity when they stick =v Then, according to question, <imgsrc="/api/proxy−image?url=https \\m×3u(i^)+2m×2u(−i^cos60∘−j^sin60∘)+3m×u(−i^cos60∘+j^sin60∘)=(m+2m+3m)v⇒3mui^−4mu2i^−4mu(23j^)−3mu2i^+3mu(23j^)=6mv⇒mui^−23mui^−23mj^=6mv⇒−21mui^−23muj^=6mv⇒v=12u(−i^−3j^)