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A 3.00 kg object is moving at 5.00 m/s EAST. It catches up to and strikes a 6.00 kg that is moving EAST at 2.00 m/s. The objects undergo an elastic collision in the EAST-WEST direction. The velocity of the 3.00 kg object after the collision is

Respuesta :

Answer:

The correct solution will be "1 m/s". The further explanation is given below.

Explanation:

The given values are:

m₁ = 3.00 kg

m₂ = 6.00 kg

v₁i = 5.00 m/s

v₂i = 2.00 m/s

Now,

⇒  [tex]v_1f = \frac{(\frac{m_{1}-m_2}{m_1+m_2} ) v1i }{(\frac{2m_2}{m_1+m_2} ) v_2i}[/tex]

On substituting the values, we get

⇒        [tex]=(\frac{3-6}{3+6} )\times 5 + (\frac{2\times 6}{3+6} )\times 2[/tex]

⇒        [tex]= 1 \ m/s[/tex]

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