Initial velocity in an inelastic collision

In summary, the drunken driver's car travelling at 1030 kg, experienced a frictional force of 276.4 N.
  • #1
EMJ
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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: MaVa + MbVb = MaVa' + MbVb'
Conservation of momentum of KE: (1/2)MaVa2+ (1/2)MbVb2 = (1/2)MaVa'2 + (1/2)MbVb'2
Frictional force = μFn

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: MaVa = V' (Ma + Mb)
I don't know how to go about this problem, and any help would be greatly appreciated. Thank you!
 
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  • #2
EMJ said:
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.
 
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  • #3
EMJ said:
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.
 
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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.
 
  • #5
EMJ said:

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: MaVa + MbVb = MaVa' + MbVb'
Conservation of momentum of KE: (1/2)MaVa2+ (1/2)MbVb2 = (1/2)MaVa'2 + (1/2)MbVb'2
Frictional force = μFn

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: MaVa = V' (Ma + Mb)
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.
 

Related to Initial velocity in an inelastic collision

What is initial velocity in an inelastic collision?

Initial velocity in an inelastic collision refers to the speed and direction at which an object is moving before it collides with another object. It is an important factor in determining the outcome of the collision.

How is initial velocity calculated in an inelastic collision?

Initial velocity can be calculated by using the formula v = (m1u1 + m2u2)/(m1 + m2), where v is the final velocity, m1 and m2 are the masses of the colliding objects, and u1 and u2 are their initial velocities.

What is the difference between initial velocity in an inelastic collision and an elastic collision?

In an elastic collision, the total kinetic energy of the system is conserved, meaning that the objects bounce off each other with no loss of energy. In an inelastic collision, some of the kinetic energy is lost, and the objects stick together after the collision. This results in a different calculation for initial velocity in an inelastic collision.

Why is initial velocity important in an inelastic collision?

Initial velocity is important because it affects the final velocity and momentum of the objects involved in the collision. It can also help determine the amount of energy lost during the collision, which is important in understanding the physical forces at play.

How does the initial velocity affect the final velocity in an inelastic collision?

The initial velocity directly affects the final velocity in an inelastic collision. The higher the initial velocity, the higher the final velocity will be, assuming all other factors remain the same. This is because initial velocity is a component in the calculation of final velocity.

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