Answer:
[tex]2\sqrt{6}[/tex] [tex]\frac{mi}{s}[/tex]
Explanation:
Assuming there is no waste of energy:
[tex]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}[/tex]
