A 1050 kg sports car is moving westbound at 15.0 m/s on a level road when it collides with a 6320 kg truck driving east on the same road at 10.0 m/s. the two vehicles remain locked together after the collision. (a) What is the velocity (magnitude and direction) of the two vehicles just after the collision? (b) At what speed should the truck have been moving so that both it and the car are stopped in the collision? (c) find the change in kinetic energy of the system of two vehicles for the situations of part (a) and part (b). for which situation is the change in kinetic energy greater in magnitude?

Mass of car: m = 1050 kg

Mass of truck: M = 6320 kg

The velocity of the car: v = 15.0 m/s to west

The velocity of the truck: V = 10.0 m/s to east

Law of conservation of momentum states that the sum of momentum of all the constituents of an isolated should remain unchanged during the collision.

Let + x be eastward. Let the common velocity with which the two vehicles will move after collision be $V_t$.

Using the law of conservation of momentum-

localid="1665051092765" $$\begin{gathered} \mathrm{MV}+\mathrm{mv}=(\mathrm{M}+\mathrm{m}){\mathrm{V}}_{\mathrm{t}} \\ \left(\right(1050\hspace{0.33em}\mathrm{kg})\times (-15.0\hspace{0.33em}\mathrm{m}/\mathrm{s}\left)\right)+\left(\right(6320\hspace{0.33em}\mathrm{kg})\times (10\hspace{0.33em}\mathrm{m}/\mathrm{s}\left)\right)=(1050\hspace{0.33em}\mathrm{kg}+6320\hspace{0.33em}\mathrm{kg})\mathrm{V} \\ \mathrm{V}=\frac{\left(\right(1050\hspace{0.33em}\mathrm{kg})\times (-15.0\hspace{0.33em}\mathrm{m}/\mathrm{s}\left)\right)+\left(\right(6320\hspace{0.33em}\mathrm{kg})\times (10\hspace{0.33em}\mathrm{m}/\mathrm{s}\left)\right)}{(1050\hspace{0.33em}\mathrm{kg}+6320\hspace{0.33em}\mathrm{kg})} \\ \mathrm{V}=6.44\hspace{0.33em}\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the final velocity is 6.44 m/s , eastward.

If both the truck and the car are stopped after collision, equation for conservation of momentum gives-

$$\begin{gathered} MV+mv=0 \\ \left(\right(6320\hspace{0.33em}kg)\times V)+(1050kg\times (-15.0\hspace{0.33em}m/s\left)\right)=0 \\ V=\frac{(1050\hspace{0.33em}kg\times (15.0\hspace{0.33em}m/s\left)\right)}{(6320\hspace{0.33em}kg)} \\ V=2.50\hspace{0.33em}m/s \end{gathered}$$

Hence, the truck would have initial speed is V = 2.50 m/s .

The change in kinetic energy during the collision in part (a).

$$\begin{gathered} ∆\mathrm{K}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}} \\ =\frac{1}{2}\left(\mathrm{M}+\mathrm{m}\right){\mathrm{v}}_{\mathrm{t}}^2-\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{MV}}^2 \\ =\frac{1}{2}\left(7370\mathrm{kg}\right){\left(6.44\mathrm{m}/\mathrm{s}\right)}^2-\frac{1}{2}\left(1050\mathrm{kg}\right){\left(10\mathrm{m}/\mathrm{s}\right)}^2 \\ =-2.81\times {10}^5\mathrm{J} \end{gathered}$$

The change in kinetic energy during the collision in part (b).

$$\begin{gathered} ∆\mathrm{K}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}} \\ =0-\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{2}{\mathrm{MV}}^2 \\ =0-\frac{1}{2}\left(1050\mathrm{kg}\right){\left(15\mathrm{m}/\mathrm{s}\right)}^2-\frac{1}{2}\left(6320\mathrm{kg}\right){\left(10\mathrm{m}/\mathrm{s}\right)}^2 \\ =-4.34\times {10}^5\mathrm{J} \end{gathered}$$

Hence, the change in kinetic energy has the greater magnitude in part (b).