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Standard XII

Physics

Perfectly Elastic Collision

Two bodies of...

Question

Two bodies of similar mass A & B collide. The velocities of A & B are $^i + 2^j$ & $2^i + 3^j$ respectively. After collision the velocity of A becomes $- 3^i - 3^j$ , then speed of B after collision is

5 m/s

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10 m/s

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20 m/s

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15 m/s

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Solution

The correct option is B 10 m/s
$M/_{A} + M/_{B} = M/ \to V_{A}^{'} + M/ - \to V_{B}^{'} 3^i + 5^j = - 3^i - 3^j +_{B} 6^i + 8^j =_{B} Speed of B = \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 m / s$

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Two bodies of similar mass A & B collide. The velocities of A & B are ^i+2^j & 2^i+3^j respectively. After collision the velocity of A becomes −3^i−3^j, then speed of B after collision is

The correct option is B 10 m/sM/→VA+M/→VB=M/→V′A+M/−→V′B3^i+5^j=−3^i−3^j+→VB6^i+8^j=→VBSpeed of B=√62+82=√36+64=√100=10 m/s

Two bodies of similar mass A & B collide. The velocities of A & B are $^i + 2^j$ & $2^i + 3^j$ respectively. After collision the velocity of A becomes $- 3^i - 3^j$ , then speed of B after collision is

The correct option is B 10 m/s
$M/_{A} + M/_{B} = M/ \to V_{A}^{'} + M/ - \to V_{B}^{'} 3^i + 5^j = - 3^i - 3^j +_{B} 6^i + 8^j =_{B} Speed of B = \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 m / s$