# How to find final velocity with given slope and mass?

Sneakatone
Because of brake failure, an automobile parked on a hill of slope 1:10 rolls 12 m downhill and strikes a parked automobile. The mass of the first automobile is 1400 kg, and the mass of the second automobile is 800 kg. Assume that the first auto- mobile rolls without friction and that the collision is elastic.

(a) What are the velocities of both automobiles immediately after the collision?

I assume the equation 1/2m1v1^2+1/2m2v2^2=1/2m1v'1^2+1/2m2v'2^2 is applied but I don't know what to do without velocity.

(b) After the collision, the first automobile continues to roll downhill, with acceleration, and the second automobile skids downhill, with deceleration. Assume that the second automobile skids with all its wheels locked, with a coefficient of sliding friction 0.90. At what time after the first collision will the automobiles have another collision, and how far from the initial collision?

barryj
Maybe you should consider conservation of momentum.

Sneakatone
you mean m1v1+m2v2=m1v'1+m2v'2 ?

barryj
That is correct. Use both the conservaqtion of energy equation (since the collision is elastoc) and the conservation of momentum.

Sneakatone
can the slope be used as a velocity?

barryj
You should use the slope and the distance traveled to find the initial velocity of the moving car.

The geometries of the problem are interesting. After the collision, I would assume the car on the slope would bounce back up the slope in the same direction but the car on the ground would travel horizontally. You might have tpo find the velocity of the parked car after the collision as if it traveled in line with the slope and then take the horizontal component.

barryj
oops, after rereading the aprioblem, the parked car is on the hill so it would travel along the hill after the collision, i.e. at the same slope.

barryj
What did you get for the 1400 kg cars velocity at the time of the collision?

Sneakatone
I still don't understand how to find velocity.

barryj
Try this:
Convert the change in potential energy of the 1400 kg car at top or ramp to kinetic energy at the point of collision. Set mgh = (1/2)mv^2 and you can solve for v. Find h from the slope and the 12m distance.

Sneakatone
12*1/10-> h=1.2
1400*9.81*1.2=.5*1400*v^2
v=4.87m/s

applying the same rule for 800 kg
v=4.85m/s

barryj
Not sure why you would use the same rule for the 800 kg car.
Now that you have the velocity you can use the equations in post 1 and 3 to solve for everything.

Sneakatone
since I don't know v2 can I use m1v1=m1v'1?

Sneakatone
I used the equation [(m1-m2)/(m1+m2)]*v1=1.32 m/s for final velocity.(1400kg)

[(2m1)/(m1+m2)]*v1=v2'
v2'=6.19 m/2 for final velocity of (800kg)

Last edited: