Question

An explosion breaks a rock into three parts, in a horizontal plane. Two of them go off at right angles to each other. The first part of mass $1\text{kg}$ moves with a speed of $12{\text{ms}}^{-1}$ and the second part of mass $2\text{kg}$ moves with $8{\text{ms}}^{-1}$ speed. If the third part files off with $4{\text{ms}}^{-1}$ speed, then its mass is-

• A. $17\text{kg}$
• B. $3\text{kg}$
• C. $5\text{kg}$
• D. $7\text{kg}$

Answer 1

The correct option is C $5\text{kg}$

According to the conversation of momentum,

$P_i=P_f$

$0=\overrightarrow{P_1}+\overrightarrow{P_2}+\overrightarrow{P_3}(\because P_i=0)$

$\therefore \overrightarrow{P_1}+\overrightarrow{P_2}=-\overrightarrow{P_3}$

$\Rightarrow$ the magnitude of $\overrightarrow{P_3}$ is equal to the magnitude of, the resultant of $\overrightarrow{P_1}$ and $\overrightarrow{P_2}$

Considering their magnitudes,

$\sqrt{(P_1{)}^2+(P_2{)}^2}=P_3$

$\sqrt{(m_1{v}_1{)}^2+(m_2{v}_2{)}^2}=m_3{v}_3$

$\sqrt{(1\times 12{)}^2+(2\times 8{)}^2}=m\left(4\right)$

$\therefore m=5\text{kg}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.