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எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

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Subjects
Science State Board
Class 10
Laws of motion
Textbook questions

28. Answer in detail VI

Exercise condition:

5 m.

State and prove the law of conservation of linear momentum.

Statement:

There is of bodies as long as acts on them.

Proof:

Consider two bodies, A and B, having masses

$m_1$ and

$m_2$ , move with an initial velocity

$u_1$ and

$u_2$ in a straight line.

Let the initial velocity of the body, A be higher than that of the body, B. i.e.,

${u_1} > {u_2}$ .

During a period of time

$t$

$second$ , they tend to have a collision. After the impact, both bodies move along the same straight line with a velocity

$v_1$ and

$v_2$ , respectively.

Force on body B due to A,

$F_{B} = m_{2}\times \left ( \frac{v_{2}-u_{2}}{t} \right )$
$F_{B} = m_{1}\times \left ( \frac{v_{1}-u_{1}}{t} \right )$
$F_{B} = m_{1}\times \left ( \frac{v_{2}-u_{2}}{t} \right )$
$F_{B} = m_{2}\times \left ( \frac{u_{2}-v_{2}}{t} \right )$

Force on body A due to B,

$F_{A} = m_{2}\times \left ( \frac{v_{2}-u_{2}}{t} \right )$
$F_{A} = m_{1}\times \left ( \frac{u_{2}-v_{2}}{t} \right )$
$F_{A} = m_{1}\times \left ( \frac{v_{1}-u_{1}}{t} \right )$
$F_{A} = m_{1}\times \left ( \frac{v_{2}-u_{2}}{t} \right )$

By

$Newton’s$

$III$

$law$

$of$

$motion$ ,

$Action force\ =\ Reaction force$

(

$m_1 \times u_1$ )

$+$

$X$

$=$

$Y$

$+$

$Z$

$X$

$=$

( ${m_1 \times u_1}^2$ )
( $m_2 \times v_2$ )
( $m_1 \times v_1$ )
( $m_2 \times u_2$ )

$Y$

$=$

( ${m_1 \times v_1}^2$ )
( $m_1 \times v_1$ )
( $m_1 \times u_1$ )
( $m_2 \times u_2$ )

$Z$

$=$

( ${m_1 \times v_2}^2$ )
( $m_1 \times u_1$ )
( $m_2 \times v_2$ )
( $m_2 \times u_2$ )

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