(a) The magnitude of the velocity of the 1000 kg car after the collision is 10.5 m/s.
(b) The direction of the velocity of the 1000 kg car after the collision is 8.2 ⁰ north west.
(c) The ratio of the kinetic energy of the two cars just after the collision to that just before the collision is 0.72.
<h3>Velocity of the 1000 kg after the collision</h3>
Apply the principle of conservation of linear momentum as follows;
<h3>Final velocity in x direction</h3>
m₁u₁ + m₂u₂ = m₁v₁x + m₂v₂x
where;
- m₁ is mass of 750 kg car
- u₁ is initial velocity of 750 kg mass
- m₂ is mass of 1000 kg car
- u₂ is initial velocity of 1000 kg mass
- v₁ is final velocity of 750 kg mass
- v₂ is final velocity of 1000 kg mass
750(0) + 1000(13) = 750(4 cos 30) + 1000v₂x
13000 = 2,598.1 + 1000v₂x
10,401.9 = 1000v₂x
v₂x = 10.4 m/s
<h3>Final velocity in y direction</h3>
m₁u₁ + m₂u₂ = m₁v₁y + m₂v₂y
750(0) + 1000(0) = 750(4 sin 30) + 1000v₂y
0 = 1500 + 1000v₂y
v₂y = -1500/1000
v₂y = -1.5 m/s
<h3>Resultant final velocity</h3>
v = √(v₂ₓ² + v₂y²)
v = √[(10.4)² + (-1.5)²]
v = 10.5 m/s
<h3>Direction of the final velocity of 1000 kg car</h3>
tanθ = v₂y/v₂ₓ
tanθ = -1.5/10.4
tanθ = -0.144
θ = arc tan(-0.144)
θ = 8.2 ⁰ north west
<h3>Kinetic energy of the cars before the collision</h3>
K.Ei = 0.5m₁u₁² + 0.5m₂u₂²
K.Ei = 0.5(750)(0)² + 0.5(1000)(13)²
K.Ei = 84,500 J
<h3>Kinetic energy of the cars after the collision</h3>
K.Ef = 0.5(750)(4)² + 0.5(1000)(10.5)²
K.Ef = 61,125 J
<h3>Ratio of the kinetic energy</h3>
K.Ef/K.Ei = 61,125/84,500
K.Ef/K.Ei = 0.72
Thus, the magnitude of the velocity of the 1000 kg car after the collision is 10.5 m/s.
The direction of the velocity of the 1000 kg car after the collision is 8.2 ⁰ north west.
The ratio of the kinetic energy of the two cars just after the collision to that just before the collision is 0.72.
Learn more about kinetic energy here: brainly.com/question/25959744
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