A particle of mass 5 kg strikes a body of mass 2.5 kg at rest elastically. The 2.5 kg mass after collision moves with a kinetic energy of 5J. The Kinetic energy of 5 kg mass before collision is

# A particle of mass 5 kg strikes a body of mass 2.5 kg at rest elastically. The 2.5 kg mass after collision moves with a kinetic energy of 5J. The Kinetic energy of 5 kg mass before collision is

1. A
0.62 J
2. B
5.625 J
3. C
7.68J
4. D
4.20 J

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### Solution:

${\text{m}}_{\text{1}}=5\text{\hspace{0.17em}}\text{kg,}\text{\hspace{0.17em}}{\text{m}}_{\text{2}}=2.5\text{\hspace{0.17em}}\text{kg,}\text{\hspace{0.17em}}{\text{u}}_{\text{2}}=\text{0}$

$\text{KE}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{2}\text{.5}\text{\hspace{0.17em}}\text{kg}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{5J}$

$\frac{1}{2}{\text{mv}}_{\text{2}}^{\text{2}}=5$

${\text{v}}_{\text{2}}^{\text{2}}=\frac{10}{2.5}$

${\text{v}}_{\text{2}}=\frac{10}{5}=2\text{\hspace{0.17em}}\text{m/s}$

${\text{v}}_{\text{2}}=\frac{\left({\text{m}}_{\text{2}}-{\text{m}}_{1}\right){\text{u}}_{\text{2}}}{\left({\text{m}}_{\text{1}}+{\text{m}}_{\text{2}}\right)}+\frac{{\text{2m}}_{\text{1}}{\text{u}}_{\text{1}}}{\left({\text{m}}_{\text{1}}+{\text{m}}_{\text{2}}\right)}$

$2=0+\frac{2\left(5\right)\left({\text{u}}_{\text{1}}\right)}{7.5}$

$2=\frac{\text{2}\left(\text{5}\right){\text{u}}_{\text{1}}}{\text{7}\text{.5}}$

${\text{u}}_{1}=\frac{7.5}{5}=1.5\text{\hspace{0.17em}}\text{m/s}$

KE of 5 kg before collision$=\frac{1}{2}{\text{mu}}_{\text{1}}^{\text{2}}$

$=\frac{1}{2}×5×{\left(1.5\right)}^{2}$

$=\frac{45}{8}$

$=5.625\text{\hspace{0.17em}}\text{J}$

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