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Question Number 18965 by Tinkutara last updated on 02/Aug/17

Two blocks of masses M and 2M are  connected to each other through a light  spring as shown in figure. If we push the  mass M with a force F which cause  acceleration a in mass M, what will the  acceleration in 2M?

$$\mathrm{Two}\:\mathrm{blocks}\:\mathrm{of}\:\mathrm{masses}\:{M}\:\mathrm{and}\:\mathrm{2}{M}\:\mathrm{are} \\ $$$$\mathrm{connected}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}\:\mathrm{through}\:\mathrm{a}\:\mathrm{light} \\ $$$$\mathrm{spring}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{figure}.\:\mathrm{If}\:\mathrm{we}\:\mathrm{push}\:\mathrm{the} \\ $$$$\mathrm{mass}\:{M}\:\mathrm{with}\:\mathrm{a}\:\mathrm{force}\:{F}\:\mathrm{which}\:\mathrm{cause} \\ $$$$\mathrm{acceleration}\:\mathrm{a}\:\mathrm{in}\:\mathrm{mass}\:{M},\:\mathrm{what}\:\mathrm{will}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{in}\:\mathrm{2}{M}? \\ $$

Commented by Tinkutara last updated on 02/Aug/17

Commented by Tinkutara last updated on 03/Aug/17

But in options it is in terms of F,Mand a.

$$\mathrm{But}\:\mathrm{in}\:\mathrm{options}\:\mathrm{it}\:\mathrm{is}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{F},{M}\mathrm{and}\:{a}. \\ $$

Commented by ajfour last updated on 02/Aug/17

initially zero .

$$\mathrm{initially}\:\mathrm{zero}\:. \\ $$

Answered by ajfour last updated on 03/Aug/17

F−kx=Ma  kx=2MA  ⇒ A=((kx)/(2M))=((F−Ma)/(2M))    A=(F/(2M))−(a/2) .     (A being acceleration of 2M).

$$\mathrm{F}−\mathrm{kx}=\mathrm{Ma} \\ $$$$\mathrm{kx}=\mathrm{2MA} \\ $$$$\Rightarrow\:\mathrm{A}=\frac{\mathrm{kx}}{\mathrm{2M}}=\frac{\mathrm{F}−\mathrm{Ma}}{\mathrm{2M}} \\ $$$$\:\:\mathrm{A}=\frac{\mathrm{F}}{\mathrm{2M}}−\frac{\mathrm{a}}{\mathrm{2}}\:.\:\:\: \\ $$$$\left(\mathrm{A}\:\mathrm{being}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{2M}\right).\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Commented by Tinkutara last updated on 03/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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