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

Answer:

$2\sqrt{6}$  $\frac{mi}{s}$

Explanation:

Assuming there is no waste of energy:

$K_{1} = K_{2}\\\frac{1}{2}m_{1}v_{1_{1}}^{2} + \frac{1}{2}m_{2}v_{2_{1}}^2 = \frac{1}{2}m_{1}v_{1_{2}}^{2} + \frac{1}{2}m_{2}v_{2_{2}}^2\\\\=> m_{1}v_{1_{1}}^{2} + m_{2}v_{2_{1}}^2 = m_{1}v_{1_{2}}^{2} + m_{2}v_{2_{2}}^2\\\\m_{1} = 2 kg, m_{2} = 1 kg, v_{1_{1}} = 4 \frac{mi}{s} , v_{2_{1}} = 0\\=> 32 = 8 + v_{2_{2}}^{2} => v_{2_{2}} = 2\sqrt{6} \frac{mi}{s}$