Solve 870 kg Sports Car Collision Speed at Impact

[sumitmanhas]

Homework Statement

A 870 kg sports car collides into the rear end of a 2500 kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.9 m before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates the speed of the sports car at impact.
Part A
What was that speed?

Homework Equations

law of conservation momentum

The Attempt at a Solution

so , we have
mivi_i + 0 = m1v1_f + m2v2_f
and since final velocities are same(inelastic collision)
m1v1_i = vf( m1 +m2 )
or m1( v1_i - vf ) = m2vf - (i)

kinetic energy is not conserved
0.5m1v1_i² = 0.5vf²( m1 + m2 )
taking the mivf² to the left and since 0.5 cancel
m1(v1_i² - vf²) = m2vf²
m1 (v1_i + vf)(v1_i - vf ) = m2vf² -(ii)

dividing (i) and (II)
we get, v1 + vf = vf
so from here do i conclude that the final velocities are different otherwise answer makes no sense and instead use subscripts for different final velocities . but final velocity should be same ?

and why is that distance and coefficient of kinetic friction given need help here..

 

[horatio89]

Homework Statement

A 870 kg sports car collides into the rear end of a 2500 kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.9 m before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates the speed of the sports car at impact.
Part A
What was that speed?

Homework Equations

law of conservation momentum

The Attempt at a Solution

so , we have
mivi_i + 0 = m1v1_f + m2v2_f
and since final velocities are same(inelastic collision)
m1v1_i = vf( m1 +m2 )
or m1( v1_i - vf ) = m2vf - (i)

(i) is better rewritten as v_i= v_f(m₁+m₂)/m₁ since the question asks for the initial speed of the sports car

kinetic energy is not conserved
0.5m1v1_i² = 0.5vf²( m1 + m2 )
taking the mivf² to the left and since 0.5 cancel
m1(v1_i² - vf²) = m2vf²
m1 (v1_i + vf)(v1_i - vf ) = m2vf² -(ii)

You stated that KE is not conserved, which is true, which means that the equivalence of KE does not hold and (ii) cannot be used.

and why is that distance and coefficient of kinetic friction given need help here..:confused

Looking back at (i), you would realize that the only unknown is v_f which is the velocity immediately after the collision. The combined vehicles came to a halt, right? We have the distance, and the final velocity of the wreakage = 0. And friction provides deceleration. So, we are looking for initial velocity. Hmm... an equation comes to mind...

 

[sumitmanhas]

so like u said , using kinematics equation vf² = vi² + 2ad
where , vf = o (final velocity of the wreckage)
vi = velocity immediately after the collision (unknown)
a = $$\mu$$mg/m ( m = total mass of two cars)
d = given

so here a would have a negative sign in front of it right , cause decelerating? and from this equation formed , the value for vi goes as vf in equation - (i) and so on n so forth u find the initial velocity of the car.

right ?
thanks for ur help n time..

 

[horatio89]

Yes, you are approaching the question correctly. a must have a negative sign, otherwise, the wreakage is accelerating!