You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with mass 80.0 kg, is given a push and slides eastward. Abigail, with mass 50.0 kg, is sent sliding northward. They collide, and after the collision Sam is moving at 37.0° north of east with a speed of 6.00 m/s and Abigail is moving at 23.0° south of east with a speed of 9.00 m/s. (a) What was the speed of each person before the collision? (b) By how much did the total kinetic energy of the two people decrease during the collision?

(a)

Given Data:

The speed of Sam after collision is $v_s=6\mathrm{m}/\mathrm{s}$

The speed of Abigail after collision is $v_a=9\mathrm{m}/\mathrm{s}$

The mass of Sam is: $m_s=80\mathrm{kg}$

The mass of Abigail is: $m_a=50\mathrm{kg}$

The angle of Sam with horizontal after collision is $\alpha =37°$

The angle of Abigail with horizontal after collision is $\beta =23°$

The conservation of momentum before and after collision of Sam and Abigail along horizontal and vertical direction is used to find the speed of Sam and Abigail before collision.

Apply the conservation of momentum along horizontal direction to find the speed of Sam before collision.

role="math" localid="1667978343816" $m_s{u}_s+m_s{u}_s\cos \alpha +m_a{v}_a\cos \beta =0$

$u_s$is the speed of Sam before collision.

role="math" localid="1667979421487" $$\begin{gathered} \left(80kg\right)u_s+\left(80kg\right)\left(6m/s\right)\left(\cos 37°\right)+\left(50kg\right)\left(9m/s\right)\left(\cos 23°\right)=0 \\ {u}_s=10m/s \end{gathered}$$

Apply the conservation of momentum along vertical direction to find the speed of Sam before collision.

$m_a{u}_a+m_s{v}_s\sin \alpha =m_a{v}_a\sin \beta$

$u_s$is the speed of Sam before collision.

$$\begin{gathered} \left(50kg\right)u_a+\left(80kg\right)\left(6m/s\right)\left(\sin 37°\right)=\left(50kg\right)\left(9m/s\right)\left(\sin 23°\right) \\ {u}_s=-2.26m/s \end{gathered}$$

Therefore, the speed of Sam and Abigail before collision are 10 m/s and 2.26 m/s .

(b)

The change in kinetic energy of the system of spheres is given as:

$∆K=\left[\frac{1}{2}{m}_s{v}_s^2+\frac{1}{2}{m}_a{v}_a^2\right]-\left[\frac{1}{2}{m}_s{v}_s^2+\frac{1}{2}{m}_a{u}_a^2\right]$

Substitute all the values in the above equation.

$$\begin{gathered} ∆K=\left[\frac{1}{2}\left(80\mathrm{kg}\right){\left(10\mathrm{m}/\mathrm{s}\right)}^2+\frac{1}{2}\left(50\mathrm{kg}\right){\left(2.26\mathrm{m}/\mathrm{s}\right)}^2\right]-\left[\frac{1}{2}\left(80\mathrm{kg}\right){\left(6\mathrm{m}/\mathrm{s}\right)}^2+\frac{1}{2}\left(50\mathrm{kg}\right){\left(9\mathrm{m}/\mathrm{s}\right)}^2+\right] \\ ∆K=663\mathrm{J} \end{gathered}$$

Therefore, the decrease in kinetic energy of both persons is 663 J .