Question

explain two bodies in contact on a horizontal surface .for 5mark quiz from 11th std​

Answer 1

Explanation:

$\bf \huge{ \underline{ \underline{ \purple{AnswEr:-}}}}$

Consider two blocks of masses $\sf m_1 \:\: and \:\: m_2 \:\:(m_1>m_2)$ are kept in contact with each other on a smooth horizontal frictionless surface

To find Acceleration:-

$\sf \: m = m_1 + m_2$

By applying Newton's 2nd law :

$\sf \vec{F} = m \vec{a}$

$\sf \:F \vec{i} = ma \vec{i}$

$\bf \: m = m_1 + m_2$

Comparing the components,

$\sf \: F = ma$

$\sf \: F = (m_1 + m_2)a$

$\sf \: a = \frac{F}{m_1 + m_2}$

$\sf \: The \: \: force \: \: exerted \: \: by \: \: the \: \: block \: \: \sf m_1 \:\: on \: m_2 \: \: due \: \: to \: \: its \: \: motion \: \: is \: \: called \: \: force \: \: of \: \: contact.( \vec{F}_{21})$

$\sf \: According \: \: to \: \: Newton's \: \: third \: \: law \: the \: block \: m_2 \: will \: exert \: \: an \: \: equal \: and \: opposite \: reaction \: force \: (\vec{F}_{12})$

Taking block m1 alone,

Refer the attachment for diagram

In x direction , applying Newton's 3rd law

$\sf \: \vec{ F } = m \vec{a}$

$\sf \: F \vec{i} + F _{12}( \vec{ - i}) = m _1a \vec{i}$

$\sf \: F \vec{i} - F _{12} \vec{i} = m _1a \vec{i}$

Comparing the components,

$\sf \: F - F _{12} = m_1a$

$\sf \: F _{12} = F - m _1a$

we know that,

$\bf \: a \: = \frac{F }{m _1 +m _2 }$

$\sf \: F _{12} = F - m _1a$

$\sf \: F _{12} = F (1 - \frac{m _1}{m _1 + m _2} )$

$F _{12} = F ( \frac{m _2}{m _1 +m _2 } )$

$\sf \vec{F _{12}} = \frac{ -F m_2 }{m _1 + m _2} \vec{i}$

Taking block m2 alone,

Refer the attachment for diagram

$\sf \vec{F} = m \vec{a}$

$\sf \: F_{21} \vec{i} = m_2a \vec{i}$

Comparing the components

$F_{21} = m_2a$

we know that,

$\bf \: a = \frac{F}{m_1 + m_2}$

$\sf \: F_{21} \frac{=m_2 F}{=m_1 +m_2 }$

$\sf \vec{F_{21}} = \frac{m_2F}{m_1 + m_2} \vec{i}$

We know that,

$\bf \: F_{12} = \frac{-m_2F}{m_1+m_2}$

$\huge \bigstar \boxed{ \mathcal { \blue{ \vec{F_{21}} = \vec{ - F_{12}}}}}$

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