Initial velocity in an inelastic collision

[EMJ]

Homework Statement

A drunken driver crashes his car into a parked car that has its brakes set. The two cars move off together (perfectly inelastic collision) for 6.0 m.
If the mass of the moving car is 1030 kg and the mass of the parked car is 1410 kg how fast was the first car traveling when it hit the parked car?
Assume that the driver of the moving car made no attempt to brake and that the coefficient of friction between the tires and the road is 0.02.

Homework Equations

Conservation of momentum: M_aV_a + M_bV_b = M_aV_a' + M_bV_b'
Conservation of momentum of KE: (1/2)M_aV_a²+ (1/2)M_bV_b² = (1/2)M_aV_a'² + (1/2)M_bV_b'²
Frictional force = μF_n

The Attempt at a Solution

I calculated the frictional force acting on the car at rest: 276.4 N
And also the frictional force acting on the whole system: 479 N
Because I'm not sure which one to use.
Since the second car is at rest and they have the same velocity following the collision, the conservation of momentum can be simplified to: M_aV_a = V' (M_a + M_b)
I don't know how to go about this problem, and any help would be greatly appreciated. Thank you!

 

[Doc Al]

I calculated the frictional force acting on the car at rest: 276.4 N
And also the frictional force acting on the whole system: 479 N
Because I'm not sure which one to use.

I would assume that the friction force only acts on the parked car since only its brakes were set.

 

[gneill]

I calculated the frictional force acting on the car at rest: 276.4 N
And also the frictional force acting on the whole system: 479 N
Because I'm not sure which one to use.

Consider that it's unlikely that the first car ends up on top of the second. Also consider the directions that the forces between the two cars are acting. Will they have any contribution to the normal force on the second car?

edit: Oops! Doc Al got there ahead of me.

 

[neilparker62]

If m₁ is the mass of the moving car and m₂ that of the stationary, then the overall energy 'budget' ½m₁ v² goes towards 'crash energy' (energy loss during a perfectly inelastic collision) and then work done against friction:

$$½m_1v^2=½m_{red}v^2+F_fΔx$$

where $m_{red}$ is the 'reduced mass' of the colliding objects: $\frac {m_1m_2}{m_1+m_2}$

Presuming that both vehicles come to a halt after the 6m.

 

[ZapperZ]

Homework Statement

A drunken driver crashes his car into a parked car that has its brakes set. The two cars move off together (perfectly inelastic collision) for 6.0 m.
If the mass of the moving car is 1030 kg and the mass of the parked car is 1410 kg how fast was the first car traveling when it hit the parked car?
Assume that the driver of the moving car made no attempt to brake and that the coefficient of friction between the tires and the road is 0.02.

Homework Equations

Conservation of momentum: M_aV_a + M_bV_b = M_aV_a' + M_bV_b'
Conservation of momentum of KE: (1/2)M_aV_a²+ (1/2)M_bV_b² = (1/2)M_aV_a'² + (1/2)M_bV_b'²
Frictional force = μF_n

The Attempt at a Solution

I calculated the frictional force acting on the car at rest: 276.4 N
And also the frictional force acting on the whole system: 479 N
Because I'm not sure which one to use.
Since the second car is at rest and they have the same velocity following the collision, the conservation of momentum can be simplified to: M_aV_a = V' (M_a + M_b)
I don't know how to go about this problem, and any help would be greatly appreciated. Thank you!

There are two issues with your setup here:

1. For an inelastic collision, KE is not conserved. So your conservation of KE equation is incorrect.

2. You have introduced one unnecessary variable. When they both stuck together after the collision, the TOTAL MASS of both vehicles move with the SAME VELOCITY. Instead, you introduced two different symbols for two different velocities for each vehicle.

The only ambiguity that I have in the problem is the statement "Assume that the driver of the moving car made no attempt to brake..." Does this mean that the driver doesn't break even after the collision? If this is true, then his wheels do not slide, and his vehicle does not contribute to the frictional force.

Zz.