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Home » A smooth sphere of mass ‘M’ moving with velocity ‘u’ directly collides elastically with another sphere of mass ‘ ‘ at rest. After collision, their final velocities are V’ and respectively. The value of is given byA)B) C)D)
A smooth sphere of mass ‘M’ moving with velocity ‘u’ directly collides elastically with another sphere of mass ‘ \mathrm{m} ‘ at rest. After collision, their final velocities are V’ and V respectively. The value of V is given by
A)\frac{2 \mathrm{u}}{1+\frac{\mathrm{M}}{\mathrm{m}}}
B) \frac{2 \mathrm{um}}{\mathrm{M}}
C)\frac{2 \mathrm{u}}{1+\frac{\mathrm{m}}{\mathrm{M}}}
D)\frac{2 \mathrm{u} \mathrm{M}}{\mathrm{m}}
Physics
A smooth sphere of mass ‘M’ moving with velocity ‘u’ directly collides elastically with another sphere of mass ‘ \mathrm{m} ‘ at rest. After collision, their final velocities are V’ and V respectively. The value of V is given by
A)\frac{2 \mathrm{u}}{1+\frac{\mathrm{M}}{\mathrm{m}}}
B) \frac{2 \mathrm{um}}{\mathrm{M}}
C)\frac{2 \mathrm{u}}{1+\frac{\mathrm{m}}{\mathrm{M}}}
D)\frac{2 \mathrm{u} \mathrm{M}}{\mathrm{m}}

Correct option is C.

Using Newton’s law of collision,
\begin{array}{l} \frac{\mathrm{V}-\mathrm{v}}{\mathrm{u}-0}=-\mathrm{e} \\ \frac{\mathrm{V}-\mathrm{v}}{\mathrm{u}-0}=-1 \quad \quad \Longrightarrow \mathrm{V}=\mathrm{v}-\mathrm{u} \quad(\because \mathrm{e}=1) \end{array}
Applying conservation of momentum just before and after the collision :
\mathrm{Mu}+0=\mathrm{MV}+\mathrm{mv}
\mathrm{Mu}=\mathrm{M}(\mathrm{v}-\mathrm{u})+\mathrm{mv}
\Longrightarrow \mathrm{v}=\frac{2 \mathrm{Mu}}{\mathrm{M}+\mathrm{m}}
\mathrm{OR} \quad \mathrm{v}=\frac{2 \mathrm{u}}{1+\frac{\mathrm{m}}{\mathrm{M}}}

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