Given,
The mass of the bumper car A is
[tex]M_A=281\text{ kg}[/tex]The mass of the bumper car B is
[tex]M_B=209\text{ kg}[/tex]The velocity of the bumper car A is
[tex]V_A=2.82\text{ m/s}[/tex]The velocity of the bumper car B is
[tex]V_B=1.72\text{ m/s}[/tex]To find: The velocity of car B after the collision.
Let v₁ and v₂ denote the velocity of car A and car B, respectively, after the collision. It sounds like both cars are initially moving in the same direction (since both have positive initial velocity).
Since momentum is conserved,
[tex]\begin{gathered} (281\text{ kg\rparen}\times(2.82\text{ m/s\rparen+\lparen209 kg\rparen}\times(1.72\text{ m/s\rparen=281 kg}\times v_1+209\text{ kg}\times v_2 \\ 1151.9\text{ kg.m/s= 281v}_1+209v_2.............(1) \end{gathered}[/tex]The kinetic energy is also conserved.
[tex]\begin{gathered} \frac{1}{2}\times281\times(2.82)^2+\frac{1}{2}\times209\times(1.72)^2=\frac{1}{2}\times281\times v_1^2+\frac{1}{2}\times209\times v_2^2 \\ 2852.93\text{ }\frac{m^2}{s^2}=281v_1^2+209v_2^2..........(2) \end{gathered}[/tex]Solve the first equation for v₁ :
[tex]v_1=\frac{1151.9-209v_2}{281}[/tex]Substitute v₁ into the second equation and solve for v₂.
[tex]\begin{gathered} 2852.93\text{ }\frac{m^2}{s^2}=281[\frac{1,151.9-209v_{2}}{281}]^2+209v_2^2 \\ v_2^2=2.96\text{ }\frac{m^2}{s^2},\text{ 8.89 }\frac{m^2}{s^2} \\ v_2=1.72\text{ m/s, 2.98 m/s} \end{gathered}[/tex]Where we ignore the first solution since it corresponds to the initial condition.
Thus, the velocity of car B after the collision is
[tex]v_2=2.98\text{ m/s}[/tex]