# Two blocks of mass 300 kg and 200 kg are moving toward eachother along a horizontal frictionless surface with velocities of and respectively. Find the final velocity of eachblock if the collision is completely elastic.

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Conservation of momentum :

200 * 100 - 300 * 50 = 200 * v1 + 300 * v2

2 v1 + 3 v2 = 50 --- (1)

Conservation of energy :

1/2 * 200 * v1² + 1/2 * 300 * v2² = 1/2 * 200 * 100² + 1/2 * 300 * 50²

2 v1² + 3 v2² = 27, 500 --- (2)

we need to solve these equations:

2 v1² + 3 (50 - 2 v1)²/3² = 27, 500

6 v1² + (2500 + 4 v1² - 200 v1) = 82,500

v1² - 20 v1 - 8, 000 = 0

v1 = 100 m/s

=> v2 = (50 - 2 v1) / 3 = - 50 m/s

200 * 100 - 300 * 50 = 200 * v1 + 300 * v2

2 v1 + 3 v2 = 50 --- (1)

Conservation of energy :

1/2 * 200 * v1² + 1/2 * 300 * v2² = 1/2 * 200 * 100² + 1/2 * 300 * 50²

2 v1² + 3 v2² = 27, 500 --- (2)

we need to solve these equations:

2 v1² + 3 (50 - 2 v1)²/3² = 27, 500

6 v1² + (2500 + 4 v1² - 200 v1) = 82,500

v1² - 20 v1 - 8, 000 = 0

v1 = 100 m/s

=> v2 = (50 - 2 v1) / 3 = - 50 m/s