
An isolated particle of mass ${\rm{m}}$moving in a horizontal plane in (xy), along the x-axis, at a certain height above the ground. It suddenly explodes into two fragments of masses $\dfrac{m}{4}$ and $\dfrac{{3m}}{4}$. An instant later, the smaller fragment is at $y = + 15{\rm{cm}}$. The large fragment at this instant is at
A.$y = - 5{\rm{cm}}$
B. $y = + 20{\rm{cm}}$
C. $y = + 5{\rm{cm}}$
D. $y = - 20{\rm{cm}}$
Answer
586.2k+ views
Hint: Centre of mass x and y coordinates does not change because the external force is equal to zero. The center of mass is that point on a body or an object where all the mass of the unevenly distributed masses of a body is concentrated.
Complete step by step answer:
A particle was moving along the x-axis before the explosion and the y component has no velocity. Therefore,
${y_{CM}} = 0$
Now, we can use here the center of mass equation and here, ${y_1}$ is the center of mass of the first part of the body from the Y axis and ${y_2}$ is the center of mass of the second part of the body from the Y axis, so, ${y_{CM}} = \dfrac{{{m_1}{y_1} + {m_2}{y_2}}}{{{m_1} + {m_2}}}$.
Here ${y_{CM}}$ is the center of mass and ${m_1}$, ${m_2}$ are the masses of the first and second part of the body.
Therefore, by simplifying the above equation it comes out to be,
$\Rightarrow \dfrac{{\left( {\dfrac{m}{4}} \right)\left( {15} \right) + \left( {\dfrac{{3m}}{4}} \right)\left( y \right)}}{{\left( {\dfrac{m}{4}} \right) + \left( {\dfrac{{3m}}{4}} \right)}} = 0$
$\Rightarrow y = - 5\;{\rm{cm}}$
Therefore, the center of mass is at -5 cm from the y-axis and the correct option is (A).
Note:
Other fragments will be at $y = 5\;{\rm{cm}}$ so that the center of mass is unaltered. The center of mass will remain at the same position before and after the origin Since all of the mass is concentrated at the center of mass and if the force is applied at the center of mass, the object will face a linear acceleration without an angular acceleration. Since there is no linear acceleration on the body, and the force is acting on the center of mass so the body will not topple about a point and there is no net torque acting on the body.
Complete step by step answer:
A particle was moving along the x-axis before the explosion and the y component has no velocity. Therefore,
${y_{CM}} = 0$
Now, we can use here the center of mass equation and here, ${y_1}$ is the center of mass of the first part of the body from the Y axis and ${y_2}$ is the center of mass of the second part of the body from the Y axis, so, ${y_{CM}} = \dfrac{{{m_1}{y_1} + {m_2}{y_2}}}{{{m_1} + {m_2}}}$.
Here ${y_{CM}}$ is the center of mass and ${m_1}$, ${m_2}$ are the masses of the first and second part of the body.
Therefore, by simplifying the above equation it comes out to be,
$\Rightarrow \dfrac{{\left( {\dfrac{m}{4}} \right)\left( {15} \right) + \left( {\dfrac{{3m}}{4}} \right)\left( y \right)}}{{\left( {\dfrac{m}{4}} \right) + \left( {\dfrac{{3m}}{4}} \right)}} = 0$
$\Rightarrow y = - 5\;{\rm{cm}}$
Therefore, the center of mass is at -5 cm from the y-axis and the correct option is (A).
Note:
Other fragments will be at $y = 5\;{\rm{cm}}$ so that the center of mass is unaltered. The center of mass will remain at the same position before and after the origin Since all of the mass is concentrated at the center of mass and if the force is applied at the center of mass, the object will face a linear acceleration without an angular acceleration. Since there is no linear acceleration on the body, and the force is acting on the center of mass so the body will not topple about a point and there is no net torque acting on the body.
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