What was the speed of the Toyota at impact in this collision?

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In a collision between a 1,000 kg Toyota and a 2,200 kg Cadillac, the Toyota's speed at impact is to be determined. The two vehicles lock bumpers and skid 2.8 meters before stopping, with a kinetic friction coefficient of 0.40. Momentum conservation principles are applied, leading to the equation m1v1 + m2v2 = (m1 + m2)v(final). The deceleration of the combined mass is calculated using the frictional force, allowing for the relationship V_0 = √(2μgd) to find the initial speed. The correct application of these formulas will yield the Toyota's speed at impact.
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A 1.0x10^3 kg Toyota collides into the rear end of a 2.2x10^3 kg Cadillac stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8m before stopping. The police officer, knowing that the coefficient of Kinectic friction between the tires and road is 0.40, calculates the speed of the Toyota at impact. What was that speed?
 
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Exactly what have you tried so far?

HINT: Momentum is conserved!
 
ok i tried using this formula:
m1v1 + m2v2 = (m1 + m2) v(final)
m1v2 + m2v2 = (m1 + m2) times the square root of 2gd (gravity and distance)
then i isolated for v1 but i don't know how to apply the coefficient of friction to the problem...
 
I think you need to assume that the two cars instantly lock up and become a single object when the collision occurs (i.e. ignore the "crumple zone" and assume the vehicles cover no distance while becoming locked together.)

The combined system decelerates at the rate \mu g (frictional force divided by mass) so you can relate the initial speed to the distance traveled before stopping with

V_0 = \sqrt {2 \mu g d}

You also know that

V_0 = \frac {m_1 v_1}{m_1+m_2}
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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