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Acceleration

A system of t...

Question

A system of two masses $m_{1}$ and $m_{2}$ connected by a string is released from rest as shown in figure. Find the $x$ and $y$ components of acceleration of COM of the system is
[All the surfaces shown in the figure are smooth]

A

(a_{c m})_{x} = {(\frac{m_{2}}{m_{1} + m_{2}})}^{2} g

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B

(a_{c m})_{x} = \frac{m_{1} m_{2} g}{(m_{1} + m_{2})^{2}}

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C

(a_{c m})_{y} = {(\frac{m_{2}}{m_{1} + m_{2}})}^{2} g

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D

(a_{c m})_{y} = \frac{m_{1} m_{2} g}{(m_{1} + m_{2})^{2}}

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Solution

The correct option is C $(a_{c m})_{y} = {(\frac{m_{2}}{m_{1} + m_{2}})}^{2} g$
Acceleration of the system is,

$a = \frac{Net pulling force}{Total mass} = \frac{m_{2} g}{m_{1} + m_{2}}$

As we know,

$(a_{c o m})_{x} = \frac{m_{1} (a_{1})_{x} + m_{2} (a_{2})_{x}}{m_{1} + m_{2}}$ and

$(a_{c o m})_{y} = \frac{m_{1} (a_{1})_{y} + m_{2} (a_{2})_{y}}{m_{1} + m_{2}}$

Since $m_{1}$ moves along $x$ -axis and $m_{2}$ moves along $y$ -axis.

Thus, for $m_{1} \to (a_{1})_{y} = 0 m_{2} \to (a_{2})_{x} = 0$

Now,

$(a_{c o m})_{x} = \frac{m_{1} a}{m_{1} + m_{2}} = \frac{m_{1} m_{2} g}{(m_{1} + m_{2})^{2}}$

$(a_{c o m})_{y} = \frac{m_{2} a}{m_{1} + m_{2}} = {(\frac{m_{2}}{m_{1} + m_{2}})}^{2} g$

 Hence, options $(B)$ and $(C)$ are the correct answer.

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