JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Conservation of Momentum

Let the initial velocity of 1 kg mass be $u$. After collision, it moves at $-2$ m/s (reversed). Let the 5 kg mass move at $v$. By momentum conservation: $1\cdot u=1\cdot (-2)+5\cdot v$, so $u=-2+5v$. By energy conservation (elastic): $\frac{1}{2}(1)u^2=\frac{1}{2}(1)(2{)}^2+\frac{1}{2}(5)v^2$, giving $u^2=4+5v^2$. Solving: $(-2+5v{)}^2=4+5v^2$ gives $4-20v+25v^2=4+5v^2$, so $20v^2=20v$, thus $v=1$ m/s. Then $u=3$ m/s. Total momentum $=1(3)=3$ kg·m/s (A correct). Momentum of 5 kg mass $=5(1)=5$ kg·m/s (B incorrect). Velocity of CM: $v_{cm}=3/6=0.5$ m/s. KE of CM $=\frac{1}{2}(6)(0.5{)}^2=0.75$ J (C correct). Total KE $=\frac{1}{2}(1)(3{)}^2=4.5$ J (D incorrect). Correct answers: (A) and (C).