A trolley of mass 360 kg is lying on a horizontal frictionless surface. A man of mass 40 kg runs with uniform speed on the trolley by a distance of 2m in 0.5 second. What is the velocity of the trolley relative to earth?

# A trolley of mass 360 kg is lying on a horizontal frictionless surface. A man of mass 40 kg runs with uniform speed on the trolley by a distance of 2m in 0.5 second. What is the velocity of the trolley relative to earth?

1. A

0.8 m/s

2. B

0.4 m/s

3. C

1.6m /s

4. D

2.2m/s

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### Solution:

From the question,

${\mathrm{V}}_{\mathrm{com}}=0$

$⇒0={\mathrm{Mv}}_{\mathrm{t}/\mathrm{e}}+{\mathrm{mv}}_{\mathrm{m}/\mathrm{e}}$

$⇒{\mathrm{v}}_{\mathrm{m}/\mathrm{t}}=\frac{2}{\frac{1}{2}}=4$

${\mathrm{v}}_{\mathrm{m}/\mathrm{t}}={\mathrm{v}}_{\mathrm{m}/\mathrm{e}}-{\mathrm{v}}_{\mathrm{t}/\mathrm{e}}$

${\mathrm{v}}_{\mathrm{m}/\mathrm{e}}={\mathrm{v}}_{\mathrm{m}/\mathrm{t}}+{\mathrm{v}}_{\mathrm{t}/\mathrm{e}}$

${\mathrm{Mv}}_{\mathrm{t}/\mathrm{e}}+\mathrm{m}\left({\mathrm{v}}_{\mathrm{m}/\mathrm{t}}+{\mathrm{v}}_{\mathrm{t}/\mathrm{e}}\right)=0$

$\left(\mathrm{M}+\mathrm{m}\right){\mathrm{v}}_{\mathrm{t}/\mathrm{e}}+{\mathrm{mv}}_{\mathrm{m}/\mathrm{t}}=0$

${\mathrm{v}}_{\mathrm{t}/\mathrm{e}}=\frac{-{\mathrm{mv}}_{\mathrm{m}/\mathrm{t}}}{\mathrm{M}+\mathrm{m}}=\frac{-40×4}{400}=-0.4\mathrm{m}/\mathrm{s}$

Hence the correct answer is 0.4m/s

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