Q85 / 120 · VITEEE 2009 · Physics
A body of mass $5 \mathrm{~kg}$ makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to $\frac{1}{10}$ th of its original velocity. Then the mass of the second body is
- Correct answerA$4.09 \mathrm{~kg}$
- B$0.5 \mathrm{~kg}$
- C$5 \mathrm{~kg}$
- D$5.09 \mathrm{~kg}$
Solution
Mass of the first body $m_{1}=5 \mathrm{~kg}$ and for elastic collision coefficient of restitution, $e=1$.
Let initially body $m_{1}$ moves with velocity $v$ after collision velocity becomes $\left(\frac{u}{10}\right)$. Let after collision velocity of $M$ block becomes $\left(v_{2}\right)$ By conservation of momentum $m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}$ or $\quad 5 u+M \times 0=5 \times \frac{u}{10}+M v_{2}$ or $\quad 5 u=\frac{u}{2}+M v_{2}$ Since, $v_{1}-v_{2}=-e\left(u_{1}-u_{2}\right)$ or $\frac{u}{10}-v_{2}=-1(u)$ or $\frac{u}{10}+u=v_{2}$ or $\frac{11 u}{10}=v_{2}$ Substituting value of $v_{2}$ in Eq. (i) from Eq. (ii) $5 u=\frac{u}{2}+M\left(\frac{11 u}{10}\right) \quad=\frac{9 \times 10}{2 \times 11}=\frac{45}{11}$ or $5-\frac{1}{2}=M\left(\frac{11}{10}\right) \Rightarrow M=4.09 \mathrm{~kg}$ $=4.09 \mathrm{~kg}$
Let initially body $m_{1}$ moves with velocity $v$ after collision velocity becomes $\left(\frac{u}{10}\right)$. Let after collision velocity of $M$ block becomes $\left(v_{2}\right)$ By conservation of momentum $m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}$ or $\quad 5 u+M \times 0=5 \times \frac{u}{10}+M v_{2}$ or $\quad 5 u=\frac{u}{2}+M v_{2}$ Since, $v_{1}-v_{2}=-e\left(u_{1}-u_{2}\right)$ or $\frac{u}{10}-v_{2}=-1(u)$ or $\frac{u}{10}+u=v_{2}$ or $\frac{11 u}{10}=v_{2}$ Substituting value of $v_{2}$ in Eq. (i) from Eq. (ii) $5 u=\frac{u}{2}+M\left(\frac{11 u}{10}\right) \quad=\frac{9 \times 10}{2 \times 11}=\frac{45}{11}$ or $5-\frac{1}{2}=M\left(\frac{11}{10}\right) \Rightarrow M=4.09 \mathrm{~kg}$ $=4.09 \mathrm{~kg}$
Hence, the correct answer is option: 'A'.