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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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