Question

A 2 kg marble moving at 4 mi./s collides into a 1 kg marble at rest. After collision, the 2 kg marble speed decreased to 2 mi./s. Calculate the velocity and speed of the 1 kg marble immediately after colliding.

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Answer 1

`2√(6)`
`(mi)/(s)`

Step-by-step explanation:

Assuming there is no waste of energy:

`K_(1) = K_(2)\\(1)/(2)m_(1)v_{1_(1)}^(2) + (1)/(2)m_(2)v_{2_(1)}^2 = (1)/(2)m_(1)v_{1_(2)}^(2) + (1)/(2)m_(2)v_{2_(2)}^2\\\\=> m_(1)v_{1_(1)}^(2) + m_(2)v_{2_(1)}^2 = m_(1)v_{1_(2)}^(2) + m_(2)v_{2_(2)}^2\\\\m_(1) = 2 kg, m_(2) = 1 kg, v_{1_(1)} = 4 (mi)/(s) , v_{2_(1)} = 0\\=> 32 = 8 + v_{2_(2)}^(2) => v_{2_(2)} = 2√(6) (mi)/(s)`