Respuesta :
Given data:
The mass of cat with skate is m=2 kg.
The speed of cat with skate is u=5 m/s.
The mass of the cake is M=3 kg.
Applying the conservation of momentum to calculate the velocity,
[tex]mu=(m+M)V[/tex]Here, V is the combined velocity of cat and cake.
Substitute the values in above equation,
[tex]\begin{gathered} (2)(5)=(2+3)V \\ V=2\text{ m/s} \end{gathered}[/tex]Thus, the final velocity immediately afterword is 2 m/s.
The initial kinetic energy will be,
[tex]\begin{gathered} KE_i=\frac{1}{2}mu^2 \\ KE_i=\frac{1}{2}(2)(5)^2 \\ KE_i=25\text{ J} \end{gathered}[/tex]The final kinetic energy will be,
[tex]\begin{gathered} KE_f=\frac{1}{2}(m+M)V^2 \\ KE_f=\frac{1}{2}(2+3)(2)^2 \\ KE_f=10\text{ J} \end{gathered}[/tex]As calculated above, the final kinetic energy is less than the initial kinetic energy.
Thus, the final kinetic energy will be decreased.
Thus, option 5 is correct.
