Two masses of 10, kg and 20, kg respectively are connected by a massless spring as shown in figure.

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4-1.Newton's Laws of Motion

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Two masses of $10\, kg$ and $20\, kg$ respectively are connected by a massless spring as shown in figure. A force of $200\, N$ acts on the $20\, kg$ mass. At the instant when the $10\, kg$ mass has an acceleration of $12\, m\, s^{-2}$, the acceleration of the $20\, kg$ mass is ...... $ms^{-2}$

A

$2$

B

$4$

C

$10$

D

$20$

Solution

If follows from the figure that the equations of motion are $:$

$200-\mathrm{f}=20 \mathrm{a}_{1}$ $\text { and } \quad f=10 a_{2}$

where $a_{1}$ and $a_{2}$ are the accelerations for $20\, kg$ and $10\, kg$ respectively.

${\text { But }} {a_{2}=12 \mathrm{ms}^{-2}}$

${\therefore \quad} {\mathrm{f}=10 \times 12=120 \mathrm{N}}$

$a_{1}=\frac{200-120}{20}=\frac{80}{20}=4 \mathrm{ms}^{-2}$

Standard 11

Physics

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