On a very muddy football field, a 110 kg linebacker tackles an 85 kg halfback. Immediately before the collision, the line backer is slipper with a velocity of 8.8 m/s north and the half-back is sliding with a velocity of 7.2 m/s east. What is the velocity (magnitude and direction) at which the two players move together immediately after the collision?

If during a collision, the total kinetic energy of the system remains conserved, then the collisions are known as elastic collisions and if the kinetic energy is not conserved, then the collision is known as Inelastic collisions.

The mass of line-backer is M = 110 kg .

The mass of half-back is m = 85 kg .

The velocity of half-back is v = 7/2 m/s due east.

The velocity of line-backer is V = 8.8 m/s

Let the common velocity of both after collision be $v_1$. Also, consider a reference frame with y-axis along north and x-axis along east.

Applying the principle of conservation of momentum along x-axis-

$$\begin{gathered} mv=(M+m)v_{1x} \\ (85\hspace{0.33em}kg)\times (7.2\hspace{0.33em}m/s)=(85\hspace{0.33em}kg+110\hspace{0.33em}kg)v_{1x} \\ {v}_{1x}=\frac{(85\hspace{0.33em}kg)\times (7.2\hspace{0.33em}m/s)}{(85\hspace{0.33em}kg+110\hspace{0.33em}kg)} \\ {v}_{1x}=3.14\hspace{0.33em}m/s \end{gathered}$$

Here, $v_{1x}$ is x component of the common velocity after collision.

Applying the principle of conservation of momentum along y-axis -

$$\begin{gathered} MV=(M+m) \\ {v}_{1y}(110\hspace{0.33em}kg)\times (8.6\hspace{0.33em}m/s)=(85\hspace{0.33em}kg+110\hspace{0.33em}kg)v_{1y} \\ {v}_{1y}=\frac{(110\hspace{0.33em}kg)\times (8.6\hspace{0.33em}m/s)}{(85\hspace{0.33em}kg+110\hspace{0.33em}kg)} \\ {v}_{1y}=4.85\hspace{0.33em}m/s \end{gathered}$$

Here, $v_{1y}$ is y component of the common velocity after collision.

The common velocity is calculated as follows;

$$\begin{gathered} v_1=\sqrt{v_{1x}^2+v_{1y}^2} \\ =\sqrt{{\left(3.14m/s\right)}^2+{\left(4.85m/s\right)}^2} \\ =5.78m/s \end{gathered}$$

The direction is calculated as follows:

$$\begin{gathered} d={\tan }^{-1}\left(\frac{V_{1y}}{V_{1x}}\right) \\ ={\tan }^{-1}\left(\frac{4.85m/s}{3.14m/s}\right) \\ =57.08° \end{gathered}$$

Hence, the direction of velocity is $57.08°$north of east and the magnitude of velocity is 5.78 m/s .