SATHEE: Laws Of Motion Question 167

Laws Of Motion Question 167

Question: Two masses $m_{1}$ and $m_{2} (m_{1} > m_{2})$ are connected by massless flexible and inextensible string passed over massless and frictionless pulley. The acceleration of centre of mass is [J&K CET 2005]

Options:

A) ${(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} g$

B) $\frac{m_{1} - m_{2}}{m_{1} + m_{2}} g$

C) $\frac{m_{1} + m_{2}}{m_{1} - m_{2}} g$

D) Zero

Show Answer

Answer:

Correct Answer: A

Solution:

Acceleration of each mass $= a = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) g$ Now acceleration of centre of mass of the system $A_{c m} = \frac{m_{1} \vec{a_{1}} + m_{1} \vec{a_{2}}}{m_{1} + m_{2}}$ As both masses move with same acceleration but in opposite direction so $\vec{a_{1}} = - \vec{a_{2}}$ = a (let) $∴ A_{c m} = \frac{m_{1} a - m_{2} a}{m_{1} + m_{2}}$

$= (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) \times (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) \times g$

$= {(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} \times g$

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Laws Of Motion Question 167

Question: Two masses m1 and m2(m1>m2) are connected by massless flexible and inextensible string passed over massless and frictionless pulley. The acceleration of centre of mass is [J&K CET 2005]

Options:

Answer:

Solution:

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Question: Two masses $m_{1}$ and $m_{2} (m_{1} > m_{2})$ are connected by massless flexible and inextensible string passed over massless and frictionless pulley. The acceleration of centre of mass is [J&K CET 2005]