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How far, in meters, will the vehicles slide after the collision?

  • 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
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  • 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.
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  • 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 truck $m_t=1593kg$

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\

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