Answer to Question #96525 in Mechanics | Relativity for Alao

Question

Two bodies A and B of equal masses moving with velocities (10km/s, 030 °) and (15km/s,090°) impigne and coalesce. Find the initial velocity with which the composite body begins to move immediately after collision

Answer 1

Let's use the conservation of the momentum (we assume that the collision was complete inelastic and there is no rotation after collision)

$${m_1}{{\vec v}_1} + {m_2}{{\vec v}_2} = ({m_1} + {m_2})\vec u$$

Choose the x-axis along the axis from which the angles in the condition of the problem is counted and y-axis is perpendicular to this direction. Then the equation for the moments will be

$$\begin{cases} {m_1}{v_1}\cos {\alpha _1} + {m_2}{v_2}\cos {\alpha _2} = ({m_1} + {m_2})u\cos \beta \\ {m_1}{v_1}\sin {\alpha _1} + {m_2}{v_2}\sin {\alpha _2} = ({m_1} + {m_2})u\sin \beta \end{cases}$$

Let's square both equations and find the sum

$${({m_1}{v_1}\sin {\alpha _1} + {m_2}{v_2}\sin {\alpha _2})^2} + {({m_1}{v_1}\cos {\alpha _1} + {m_2}{v_2}\cos {\alpha _2})^2} = {({m_1} + {m_2})^2}u^2$$

Expand the brackets and use that ${\sin ^2}{\alpha _1} + {\cos ^2}{\alpha _1} = 1$ and $\cos {\alpha _1}\cos {\alpha _2} + \sin {\alpha _1}\sin {\alpha _2} = \cos ({\alpha _2} - {\alpha _1})$ we get

$${({m_1}{v_1})^2} + {({m_2}{v_2})^2} + 2{m_1}{m_2}{v_1}{v_2}\cos ({\alpha _2} - {\alpha _1}) = {({m_1} + {m_2})^2}{u^2}$$

Thus

$$u = {{\sqrt {{{({m_1}{v_1})}^2} + {{({m_2}{v_2})}^2} + 2{m_1}{m_2}{v_1}{v_2}\cos ({\alpha _2} - {\alpha _1})} } \over {{m_1} + {m_2}}}$$

Now return to the system and divide the second equation by the first, we get

$$\tan \beta = {{{m_1}{v_1}\sin {\alpha _1} + {m_2}{v_2}\sin {\alpha _2}} \over {{m_1}{v_1}\cos {\alpha _1} + {m_2}{v_2}\cos {\alpha _2}}}$$

These two formulas are solving our problem. Let's use that is our case ${m_1} = {m_2} = m$

$$u = {1 \over 2}\sqrt {v_1^2 + v_2^2 + 2{v_1}{v_2}\cos ({\alpha _2} - {\alpha _1})}$$ $$\beta = \arctan ({{{v_1}\sin {\alpha _1} + {v_2}\sin {\alpha _2}} \over {{v_1}\cos {\alpha _1} + {v_2}\cos {\alpha _2}}})$$

And do all calculations

$$u = {1 \over 2}\sqrt {{{10}^2} + {{15}^2} + 2 \cdot 10 \cdot 15 \cdot {1 \over 2}} [{{{\rm{km}}} \over {\rm{s}}}] = {{5\sqrt {19} } \over 2}[{{{\rm{km}}} \over {\rm{s}}}]$$ $$\beta = \arctan ({{10 \cdot {1 \over 2} + 15 \cdot 1} \over {10 \cdot {{\sqrt 3 } \over 2} + 15 \cdot 0}}) = \arctan ({4 \over {\sqrt 3 }})$$

This is the magnitude and the direction (angle with the x-axis) of the composite body after collision. We can find the velosity (as the vector). Use that $\vec u = (u\cos \beta ,u\sin \beta )$ and $\sin \arctan x = {x \over {\sqrt {1 + {x^2}} }}$ and $\cos \arctan x = {1 \over {\sqrt {1 + {x^2}} }}$ we get

$$\vec u = ({{5\sqrt {19} } \over 2} \cdot {{\sqrt 3 } \over {\sqrt {19} }},{{5\sqrt {19} } \over 2} \cdot {4 \over {\sqrt {19} }})[{{{\rm{km}}} \over {\rm{s}}}] = ({{5\sqrt 3 } \over 2},10)[{{{\rm{km}}} \over {\rm{s}}}]$$