**A car with mass mc=1074kg is traveling west through an intersection at a magnitude of veloctiy of vc=15m/s when a truck of mass mt=1593 kg traveling south at vt=10.8 m/s fails to yield and collides with the car. The vehicles become stuck together and slide on the asphalt, which has a coefficiet of friction of mk=0.5****With the variables mentioned in the above problem and the unit vectors i and j, write the equation which defines the velocity of both car and truck being stuck together after the accident.****What distance $(m)$ will both vehicles slide being stuck together after the accident?**

The aim of the question is to find the equation representing the **velocity of the system** (car and truck stuck together) and the **distance being traveled** by them in that state after the collision.

The basic concept behind the solution is $Law$ $of$ $Conservation$ $of$ $Momentum$. The $Law$ $of$ $Conservation$ $of$ $Momentum$ states that the total **momentum** $p$ of an isolated system will always remain the same.

Consider the collision of $2$ bodies having masses $m_1$ and $m_2$ with initial velocities $u_1$ and $u_2$ along straight lines, respectively. After collision, they acquire velocities $v_1$ and $v_2$ in the same direction, so **total momentum** before and after collision is defined as:

\[p_i=m_1u_1+m_2u_2\]

\[p_f=m_1v_1+m_2v_2\]

In the absence of any external force on the system:

\[p_i=p_f\]

\[m_1u_1+m_2u_2=m_1v_1+m_2v_2\]

## Expert Answer

Given that:

**Mass of the car** $m_c=1074kg$

**Velocity of the car** $v_c=15\dfrac{m}{s}(west)=-15i\dfrac{m}{s}\ (east)$ by considering east as $+ve$ $x$ direction or $+ve$ $i$

**Mass of the truc**k $m_t=1593kg$

**Velocity of the truck** $v_t=10.8\dfrac{m}{s}(south)=-15i\dfrac{m}{s}\ (north)$ by considering east as $+ve$ $y$ direction or $+ve$ $j$

**Final Velocity** of both Car and Truck stuck together $v_f=?$

**Distance **Traveled after Collision $D=?$

**Part A**

By considering the $Law$ $of$ $Conservation$ $of$ $Momentum$:

\[m_cv_c+m_tv_t=m_cv_f+m_tv_f\]

By writing the equation in terms of $v_f$:

\[m_cv_c+m_tv_t={(m}_c+m_t)v_f\]

\[v_f=\frac{m_cv_c+m_tv_t}{{(m}_c+m_t)}\]

By substituting the given values:

\[v_f=\frac{{1074kg\times(-15i)}+{1593kg\times(-10.8j)}}{(1074kg+1593kg)}\]

\[v_f=v_i+v_j=-6.04i-6.45j\]

**Part B**

The **absolute value of the velocity** of both vehicles stuck together is:

\[v_f=\sqrt{{v_i}^2+{v_j}^2}\]

\[v_f=\sqrt{{(-6.04)}^2+{(-6.45)}^2}\]

\[v_f=8.836\dfrac{m}{s}\]

After the collision, the **Kinetic energy** of both vehicles is combined against the friction force of the asphalt. The **friction force** is represented as follows:

\[F_f=\mu_k(m_c+m_t)g\]

\[F_f=0.5(1074kg+1593kg)\times9.81\frac{m}{s^2}\]

\[F_f=13,081.635\ kg\frac{m}{s^2}=13,081.635N\]

**Kinetic Energy** and its relationship with **Friction Force** $F_f$ is represented as follows:

\[K.E.=\frac{1}{2}(m_c+m_t){v_f}^2=F_f\ .D\]

\[D=\frac{1}{2}(m_c+m_t){v_f}^2\times\frac{1}{F_f}\]

\[D=\frac{(1074kg+1593kg)\times({8.836\dfrac{m}{s})}^2}{2}\times\dfrac{1}{13081.635N}=7.958m\ \]

## Numerical Result

The **Final Velocity** of both Car and Truck stuck together is:

\[v_f=-6.04i-6.45j\]

**Distance** traveled by both car and truck after the collision is:

\[D=7.958m\]

## Example

A car with a **velocity** of $v_c=9.5\dfrac{m}{s}$ and a **mass** $m_c=1225kg$ is being driven towards the west. A truck, which is moving south with a **velocity** $v_t=8.6\dfrac{m}{s}$ and a **mass** of $m_t=1654kg$, crashes with the car. Both vehicles slide on the asphalt while stuck to each other.

With the **unit vectors** $i$ and $j$, write the **equation of velocity** of both car and truck being stuck together after the collision.

**Solution**

By considering the $Law$ $of$ $Conservation$ $of$ $Momentum$ along the direction $i$ and $j$, we can write:

\[m_cv_c+m_tv_t=m_cv_f+m_tv_f\]

\[v_f=\frac{m_cv_c+m_tv_t}{{(m}_c+m_t)}\]

\[v_f=\frac{{1225kg\times(-9.5i)}+{1654kg\times(-8.6j)}}{(1225kg+1654kg)}\]

\[v_f=-4.04i-4.94j\