Respuesta :
Answer:
[tex]Q = 2000 J[/tex]
Explanation:
As we know that the 6 kg object was at rest initially
So here since net force on the system is zero
so the momentum of the system will always remains conserved
so we can say
[tex]0 = P_1 + P_2 + P_3[/tex]
now we know that
[tex]P_1 = P_2 = P[/tex]
and the angle between the two objects is 60 degree
so we can say
[tex]\vec P_1 + \vec P_2 = \sqrt{P_1^2 + P_2^2 + 2P_1 P_2cos60}[/tex]
[tex]\vec P_1 + \vec P_2 = \sqrt{P^2 + P^2 + 2P^2(0.5)} = \sqrt3 P[/tex]
now we can say that the speed of the third mass will be
[tex]v_3 = \sqrt 3 (20) m/s[/tex]
now the total kinetic energy released in this system is given as
[tex]Q = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 + \frac{1}{2}mv_3^2[/tex]
[tex]Q = \frac{1}{2}(2)(20^2) + \frac{1}{2}(2)(20^2) + \frac{1}{2}(2)(20\sqrt3)^2[/tex]
[tex]Q = 2000 J[/tex]
