91Ó°ÊÓ

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

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-

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

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

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

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

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

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

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