Answer:
The velocity of the composite body is 0.99m/s 63.43° west of north.
Explanation:
Here the law of conservation of energy says that
[tex](1).\: \: m_1v_1cos(0)+m_2v_2cos(90^o)=(m_1+m_2)v_f cos(\theta)[/tex]
[tex](2).\: \: m_1v_1sin(0)+m_2v_2sin(90^o)=(m_1+m_2)v_f sin(\theta)[/tex]
where [tex]v_f[/tex] is the final velocity if the composite body, and [tex]\theta[/tex] is measured from west of north.
Putting in numbers and simplifying the above equation we get:
[tex](3).\: \: m_1v_1=(m_1+m_2)v_f cos(\theta)[/tex]
[tex](4).\: \: m_2v_2=(m_1+m_2)v_f sin(\theta)[/tex]
dividing equation (4) by (3) gives
[tex]\dfrac{ m_2v_2=(m_1+m_2)v_f sin(\theta)}{ m_1v_1=(m_1+m_2)v_f cos(\theta)}[/tex]
[tex]\dfrac{m_2v_2}{m_1v_1} = \dfrac{sin(\theta)}{ cos(\theta)}[/tex]
[tex](5).\: \: tan(\theta) = \dfrac{m_2v_2}{m_1v_1}[/tex]
putting in [tex]m_1 = 0.5kg[/tex], [tex]v_1 = 4m/s[/tex], [tex]m_2 = 4kg[/tex], and [tex]v_2 = 1m/s[/tex] we get:
[tex]tan(\theta) = \dfrac{(4kg)(1m/s)}{(0.5kg)(4m/s)}[/tex]
[tex]\boxed{\theta = 63.43^o}[/tex]
Thus, the final velocity [tex]v_f[/tex] we get from equation (3) is:
[tex]v_f=\dfrac{m_1v_1}{(m_1+m_2)cos(\theta)}[/tex]
[tex]v_f=\dfrac{(0.5kg)(4m/s)}{(4kg+0.5kg)cos(63.43^o)}[/tex]
[tex]\boxed{v_f = 0.99m/s}[/tex]
Thus, the velocity of the composite body is 0.99m/s 63.43° west of north.
