Explanation:
Applying the law of conservation of momentum,
Momentum before collision = Momentum after collision.
[tex]m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}[/tex]
[tex]1810 \mathrm{~kg} \times 16 \frac{\mathrm{m}}{\mathrm{s}}+0=1810[/tex] [tex]\mathrm{~kg} \times \mathrm{v}_{1}+673 \times v_{2}[/tex]
[tex]28960=1810 \mathrm{~kg} \times \mathrm{v}_{1}+673 \times v_{2}[/tex]
For an elastic collision,
Relative velocity of approach = Relative velocity of separation.
[tex]16 \frac{\mathrm{m}}{\mathrm{s}}=\mathrm{v}_{2}-\mathrm{v}_{1}[/tex]
[tex]\mathrm{v}_{2}=v_{1}+16 \quad \ldots \ldots \ldots \ldots(\mathrm{b})[/tex]
Using the value of [tex]v_{2}[/tex] in equation (a)
[tex]28960 &=1810 \mathrm{v}_{1}+673\left(\mathrm{v}_{1}+16\right)[/tex]
[tex]=1810 \mathrm{y}+673 \mathrm{v}_{1}+10768[/tex]
[tex]28960-10768 &=2483 \mathrm{v}_{1}[/tex]
[tex]18192 &=2483 \mathrm{v}_{1}[/tex]
[tex]v_{1} &=7.33 \frac{\mathrm{m}}{\mathrm{s}}[/tex]
Using the value of $v_{1}$ in equation (b)
[tex]v_{2} &=7.33+16[/tex]
[tex]=23.33 \frac{\mathrm{m}}{\mathrm{s}}[/tex]
Therefore the speed of truck after collision was found to be [tex]7.33 \frac{\mathrm{m}}{\mathrm{s}}[/tex]
And the speed of car after collision was found to be [tex]23.33 \frac{\mathrm{m}}{\mathrm{s}}[/tex]
