Perpendicular 2D collision

Problem

Two identical balls travel perpendicular to each other: the first with $\vv{v}_1 = (3;\, 0)\;\text{m/s}$, the second with $\vv{v}_2 = (0;\, 4)\;\text{m/s}$. After the collision they stay joined. Calculate the final velocity (magnitude and direction).

Solution

Perfectly inelastic collision (they stay joined): total momentum is conserved, and it is a vector. The two balls have equal mass $m$; after the collision the joined mass is $2m$. Work separately on the $x$ and $y$ components. $x$ component (only ball 1 has velocity along $x$): $m\cdot 3 + m\cdot 0 = 2m\,v_{fx} \;\Rightarrow\; v_{fx} = \frac{3}{2} = 1{,}5\ \mathrm{m/s}$ $y$ component (only ball 2 has velocity along $y$): $m\cdot 0 + m\cdot 4 = 2m\,v_{fy} \;\Rightarrow\; v_{fy} = \frac{4}{2} = 2{,}0\ \mathrm{m/s}$ So the final velocity vector is: $\vv{v}_f = (1{,}5;\, 2{,}0)\ \mathrm{m/s}$ Magnitude: $\norm{\vv{v}_f} = \sqrt{v_{fx}^2 + v_{fy}^2} = \sqrt{1{,}5^2 + 2{,}0^2} = \sqrt{6{,}25} = 2{,}5\ \mathrm{m/s}$ Direction (angle relative to the $x$ axis): $\theta = \arctan\frac{v_{fy}}{v_{fx}} = \arctan\frac{2{,}0}{1{,}5} \approx 53^\circ$ The joined body moves at the velocity of the centre of mass, diagonally between the two initial directions. $\ev{\vv{v}_f = (1{,}5;\, 2{,}0)\ \mathrm{m/s},\quad \norm{\vv{v}_f} = 2{,}5\ \mathrm{m/s},\quad \theta \approx 53^\circ}$

Links

Topics: Momentum and collisions Concepts: Inelastic collision · Conservation of momentum · Centre of mass Skills: Conservation of momentum