Answer:
The final speed of cart with mass 0.06 kg is 2.7 m/s and the final speed of the cart of mass 0.09 kg is 4.2 m/s.
Explanation:
Initial moumentum of cart with mass 0.06 kg is
[tex]p_1=0.06\times 4.5=0.27kgm/s[/tex]
Initial moumentum of cart with mass 0.06 kg is
[tex]p_2=0.09\times 3=0.27kgm/s[/tex]
Thus initial moumentum of system is
[tex]p_1+p_2=2\times 0.27=0.54kgm/s[/tex]
Now let the final velocity of 0.06 kg cart be [tex]v_{f1}[/tex] and that of 0.09 kg cart be [tex]v_{f2}[/tex]
Thus final moumentum of system is
[tex]0.06\times v_{f1}+0.09\times v_{f2}[/tex]
Equating initial and final values we get
[tex]0.06\times v_{f1}+0.09\times v_{f2}=0.54........(i)[/tex]
Similarly conserving the kinetic energies of the 2 carts we get
[tex]\frac{1}{2}\times 0.06\times 4.5^{2}+\frac{1}{2}\times 0.09\times 3^{2}=\frac{1}{2}\times 0.06\times (v_{1f})^{2}+\frac{1}{2}\times 0.09\times (v_{2f})^{2}\\\\0.06\times 4.5^{2}+0.09\times 3^{2}=0.06\times (v_{2f})}^{2}+0.09\times (v_{2f})^{2}\\\\0.06\times (v_{2f})}^{2}+0.09\times (v_{2f})^{2}=2.025.......(ii)[/tex]
Solving equation i and ii we get
Substituting value of [tex]v_{2f}[/tex] from i and using it in equation ii we get
[tex]0.06v_{1f}^2+\frac{0.06^2}{0.09}v_{1f}^{2}-2\times \frac{0.54\times 0.06}{0.09}v_{1f}+\frac{0.54^2}{0.09}=2.025[/tex]
Solving for [tex]v_{1f}[/tex] we get
[tex]v_{1f}=2.7m/s[/tex]
Similarly
[tex]v_{2f}=\frac{0.54-0.06\times 2.7}{0.09}=4.2m/s[/tex]
