The mass of the red car is $m_{1} = 950 mathrm{~kg}$. The initial speed of the red car is $v_{1,i} = 16 mathrm{~m/s}$ in the positive $X$ direction. The mass of the blue minvan is $1300 mathrm{~kg}$. The initial speed of the blue minivan is $v_{2,i} = 21 mathrm{~m/s}$ in the positive $Y$ direction.

As the problem mentions, the initial speed of the red car is greater than $16 mathrm{~m/s}$. According to the conservation law of momentum. The total momentum of the system must be zero. So, if the initial momentums increases, the final momentum should be decreasing. As the textbook mentions, the definition of the momentum assigns that the momentum is the mass times velocity.

$$
begin{align*}
vec{p} &= m ~ vec{v} \
end{align*}
$$

So, any increase in the initial velocity of the red car leads to an increase in the initial momentum of the red car. Since the car and the minivan stick together after the collision. So, after the collision, the two cars move toward the right with an angle less than $60^{circ}$.

The red car of the system will add momentum to the right side in the x direction and the blue minivan adds momentum in the positive y direction. The momentums for both are the same before and after collision we will use this principle to find the components of the system’s momentum after cars collide and use this to determine the speeds of the car before they collide.

$textbf{Solution:}$
Firstly Find the momentum in the x direction

$$
p_{x,i}=p_{x,f}=p_{f}costheta=(m_1+m_2)v_fcostheta
$$

$$
=(950+1300kg)(12.5m/s)cos 42^{0}
$$

$$
p_{x,i}=20900kg.m/s
$$

Secondly the initial speed of the red car is determined

$$
p_{x,i}=m_1 v_{1,i}=20900kg.m/s
$$

$$
v_{x,i}=frac{20900kg.m/s}{m_1}=frac{20900}{950kg}
$$

$$
boxed{bf color{#4257b2}v_{1,i}=22m/s}
$$

Again the momentum in the y direction for the minivan is to be found

$$
p_{y,i}=p_{y,f}=p_{f}sintheta=(m_1+m_2)v_fsintheta
$$

$$
=(950+1300kg)(12.5m/s)sin 42^{0}
$$

$$
p_{y,i}=18800kg.m/s
$$

Now for minivan, the initial speed is determined

$$
p_{y,i}=m_2 v_{2,i}=18800kg.m/s
$$

$$
v_{y,i}=frac{18800kg.m/s}{m_2}=frac{18800}{1300kg}
$$

$$
boxed{bf color{#4257b2}v_{2,i}=14m/s}
$$

$$
v_{2,i}=14m/s
$$

The overall momentum of the players after the collision will be same as their momentum before collision. With this information we find their momentum after collision and use combined mass to find the velocity after collision.

$textbf{Solution:}$
Find momentum in x direction

$$
p_{x,i}=p_{x,f}=m_2 v_{x,2}=(96kg)(-2.4m/s)
$$

$$
p_{x,i}=-230.4 kg.m/s
$$

Find momentum in y direction

$$
p_{y,i}=p_{y,f}=m_1 v_{x,1}=(82kg)(3.1m/s)
$$

$$
p_{y,f}=-254.2 kg.m/s
$$

Now the magnitude of the final momentums is to be calculated

$$
p_f=sqrt{p^2_{x,f}+p^2_{y,f}}
$$

$$
P-f=sqrt{(-230.4kg.m/s)+(254.2kg.m/s)}
$$

$$
p_f=343.1 kg.m/s
$$

Now divide by the total mass to find the speed

$$
v_f=frac{p_f}{m_1+m_2}=frac{343.1kg.m/s}{82+96kg}
$$

$$
boxed{bf color{#4257b2}v_f=1.9m/s}
$$