Answer:
4.72 m/s
Explanation:
Since the collision is perfectly elastic, both momentum and energy is conserved
Momenum: [tex] mv_1 = mv_2 + MV[/tex]
Kinetic energy: [tex]0.5mv_1^2 = 0.5mv_2^2 + 0.5MV^2[/tex]
where m = 3.5 kg is the mass of the small moving object at speed v1 = 13 m/s. M = 7 kg is the mass of the big collided object with speed V post-collision. v2 is the speed of the small object post-collision.
Momentum: [tex]3.5*13 = 3.5v_2 + 7V[/tex]
[tex]13 = v_2 + 2V[/tex]
[tex]v_2 = 13 - 2V[/tex]
Energy: [tex]3.5*13^2 = 3.5v_2^2 + 7V^2[/tex]
[tex]169 = v_2^2 + 2V^2[/tex]
We can substitute [tex]v_2 = 13 - 2V[/tex]
[tex]169 = (13 - 2V)^2 + 7V^2[/tex]
[tex]169 = 169 - 52V + 4V^2 + 7V^2[/tex]
[tex]V(11V - 52) = 0[/tex]
[tex]11V - 52 = 0[/tex]
[tex]V = 52/11 = 4.72 m/s[/tex]
