# Potential/Kinetic Energy Problem need help

[SOLVED] Potential/Kinetic Energy Problem need help ASAP

A small sports car and a pickup truck start coasting down a 10.0 m hill together, side by side. Assuming no friction, what is the velocity of each vehicle at the bottom of the hill? Submit answer in m/s; sports car, then truck.

I don't know how to figure this one out, I know it deals with conservation of energy but I don't know how to calculate it when all I have been given is one variable.

Working at a firing range you are tasked to determine the muzzle speed of a bullet as it exits an experimental rifle.
You set up a ballistic pendulum in order to conduct your experiment. The pendulum is made of a block of wood suspended from a set of strings. You fire the bullet into the stationary block and measure the speed of the combined bullet and block since the bullet lodges inside the block. You measure the mass of the bullet to be 4.20 g, the mass of the block to be 500. g, and the speed of the combined bullet and block to be 6.75 m/s. Calculate the speed of the bullet (muzzle speed) and give your answer in m/s.

I am completely lost on this one. I am taking an online class and my instructor is taking forever answering a couple of questions I have about these problems. I just need some help with getting pointed into the right direction.

Thanks

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Bump...I need help

I feel like I'm fishing.

You should.

For the first problem, do you feel like giving some more information, like hill's base or angle?

You should.

For the first problem, do you feel like giving some more information, like hill's base or angle?
That is all that is given. I even asked my instructor how can I solve for a problem when all I know is one variable. He went on to say basically PE=x at the top of the hill and KE=0 therefore solve for PE and you will find your answer. But I still don't know how I am to solve it, with only one variable maybe if I had the mass of the vehicles.

assume energy conserved.
potential energy = kinectic energy
mgh = 0.5mv^2
gh = 0.5v^2
(9.81)(10) = 0.5v^2
v=14m/s

The problem is I asked him the speed of a falling object free falling is 9.81 m/s but these are not free falling so I didn't know where I could get a second variable from. He could have told me yes use 9.81 m/s.

For the second one.
The Conservation of Momentum.
M1= mass of bullet
V1=velocity of bullet
M2=mass of block
V2=velocity of block
Vf=final velocity of the system
M1V1+M2V2=Vf(M1+M2)
converted grams to kg
(.0042)(V1)+(.5)(0)=6.75(.0042+.5)
(V1).(0042)=3.4033
V1=810.321 m/s