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A particle of mass 3 m is projected from the ground at some angle with horizontal . The horizontal range R. At the highest point of its path it breaks into two pieces m and 2 m. The smaller mass comes to rest and larger mass finally at a distance x from the point of projection where x is equal to :

Option: 1

2 \mathrm{R}


Option: 2

\frac{5 R}{4}


Option: 3

3 R


Option: 4

\frac{7 R}{3}


Answers (1)

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Since the breaking of the particle does not involve any external forces, the motion of the center of mass remains undisturbed. Consequently, the final position of the center of mass will be at a distance R (Range) from the point of projection

\begin{gathered} 2 \mathrm{~m}(\mathrm{x}-\mathrm{R})=\frac{(m \times R)}{2} \\ \mathrm{x}=\frac{5 R}{4} \end{gathered}

Posted by

Devendra Khairwa

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