Answer:
6.16 m/s
Step-by-step explanation:
m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final
3.0 kg * 6.0 m/s) + (2.0 kg * 0 m/s) = (3.0 kg * v1_final) + (2.0 kg * (4.0 m/s * cosθ))
18 kg·m/s = 3.0 kg * v1_final + 8.0 kg * cosθ
let's consider the conservation of kinetic energy
(1/2) * m1 * (v1_initial)^2 + (1/2) * m2 * (v2_initial)^2
The total kinetic energy
(1/2) * m1 * (v1_final)^2 + (1/2) * m2 * (v2_final)^2
Since the 2.0 kg ball starts from rest
(1/2) * 3.0 kg * (6.0 m/s)^2 = 54 J
After the collision, the 2.0 kg ball moves with a velocity of 4.0 m/s, and the 3.0 kg ball moves with a velocity v1_final.
Using conservation of kinetic energy, i have:
adapt at the formula
(1/2) * 3.0 kg * (v1_final)^2 + (1/2) * 2.0 kg * (4.0 m/s)^2 = 54 J
(1/2) * 3.0 kg * (v1_final)^2 + 8 J = 54 J
(1/2) * 3.0 kg * (v1_final)^2 = 46 J
1.5 kg * (v1_final)^2 = 46 J
(v1_final)^2 = 46 J / 1.5 kg
v1_final = √(46 J / 1.5 kg)
v1_final ≈ 6.16 m/s
, the velocity of the 3.0 kg ball after the collision is approximately 6.16 m/s.
I think so I am right if mi school work