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2 years ago

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A 750 kg car is stalled on an icy road during a snowstorm. A 1000 kg car traveling eastbound at 13 m/s collides with the rear of

the stalled car. After being hit, the 750 kg car slides on the ice at 4 m/s in a direction 30 ∘ north of east.
A) What is the magnitude of the velocity of the 1000 kg car after the collision?
B) What is the direction of the velocity of the 1000 kg car after the collision?
C) Calculate the ratio of the kinetic energy of the two cars just after the collision to that just before the collision. (You may ignore the effects of friction during the collision.)

1 answer:

scZoUnD [109]2 years ago

3 0

(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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