A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed v. What is the mass of other body and the speed of the center of mass of two bodies ?

1.0 k g a n d \frac{2}{3} v

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1.2 k g a n d \frac{5}{8}

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1.4 k g a n d \frac{10}{17} v

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1.5 k g a n d \frac{4}{7} v

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Solution

The correct option is C $1.2 k g a n d \frac{5}{8}$
$m_{1} = 2 k g$
$u_{1} = 4 m / s$
$v_{1} = \frac{u_{1}}{4} = \frac{4}{4} = 1 m / s$
$u_{2} = 0$
since nothing is given so we would take it as elastic collision.
hence conservation of momentum is given by:
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$2 \times 4 + 0 = 2 \times 1 + m_{2} v_{2}$
$m_{2} v_{2} = 6$
according to the conservation kinetic energy:
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$
$2 \times 16 + 0 = 2 \times 1 + m_{2} v_{2}^{2}$
$m_{2} v_{2}^{2} = 30$
$\frac{m_{2} v_{2}^{2}}{m_{2} v_{2}} = \frac{30}{6}$
$v_{2} = 5 m / s$
$m_{2} = 1.2 k g$

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