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Laws of Motion

Section: Physics

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Question : 20 of 20

Marks: +1, -0

Two masses

m_1

m_2

are commected by a light string passing over smooth pulley. When set free

m_1

moves downwards by

3 m

3 s

\; \frac{m_1}{m_2}

( g=10 \mathrm{ms}^{-2} )

[AP EAPCET 18 May 2024 Shift 1]

$\frac{9}{7}$
$\frac{8}{7}$
$\frac{10}{7}$
$\frac{15}{13}$

Solution:

Given,

\; s=3\,\mathrm{m}, t=3\,\mathrm{s}, u=0(\; \text{ set free }\;)

\; s=u t+\; \frac{1}{2} a t^2

\; 3=0+\; \frac{1}{2} a \times 3^2

\; a=\; \frac{2}{3} \mathrm{m/s}^2

Applying the result of Newton 2nd law,

F=m a

So,

\; m_1 g-T=m_1 a

\; T-m_2 g=m_2 a

Adding Eq. (i) and Eq. (ii),

g (m_1-m_2) \; = a (m_1+m_2)

10 (m_1-m_2) \; = \; \frac{2}{3} (m_1+m_2)

15 m_1-15 m_2 \; = m_1+m_2

14 m_1 \; = 16 m_2

7 m_1 \; = 8 m_2

\; \frac{m_1}{m_2} \; = \; \frac{8}{7}

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