Two particles A and B strike each other as shown. mA = 2 kg and mB = 1.2 kg. Jus

Question

Two particles A and B strike each other as shown. mA = 2 kg and mB = 1.2 kg. Just before the collision nu A = (5i + 5j) m/s and nu B = 0 m/s. e = 1.0. What are their velocities after the collision?

Explanation / Answer

Given that       e = 1     ==> vB' - vA' = - ( vB - vA)     ==> vB' - vA' = vA = (5i + 5j)     ==> vB' = (5i + 5j) + vA' According to conservation of momentum we have    mAvA + 0 = mAvA' +mBvB' ==> 2 (5i + 5j) = 2vA' + 1.2vB'            10i + 10j = 2vA' + 1.2 * [(5i + 5j) + vA']            10i + 10j = 3.2vA' + 6i + 6j ==> 3.2vA' = 4 i + 4j ==> vA' = 1.25 i + 1.25 j          vB' = (5i + 5j) +1.25 i + 1.25 j = 6.25 i + 6.25 j

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