# A-Level Physics Problem - Velocity

1. Dec 23, 2013

### elDuderino81

1. A Level Physics - finding resultant velocity at bottom of a frictionless incline, with no angle provided?

2. Hi, Im currently working through a problem set and I'm being asked to find the speed of a cart (roller-coaster) after it has travelled down a slope (from points A to B). The trouble is, I've not been given any angles other than the right angle, and no information regarding distance travelled.

The only information I have is the mass of the cart (Mc), the hight from which it is released from rest (hA), and I have been asked to assume acceleration due to gravity as 10ms-1.

Mc=6000N
hA=50m
g=10ms-1

3. Am I being a bit dim here? The way I see it, without the angle of inclination I wont be able to find the resultant velocity right? Regardless of the inclination being frictionless, there would still be a reaction force that would effect the velocity, but without the angle I can't resolve that either? I'm a bit stumped by this.

Any assistance anyone could lend regarding this would be very much appreciated.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 23, 2013

### Staff: Mentor

Perhaps the final speed does not depend on the angle of the slope.

Hint: What's conserved?

3. Dec 23, 2013

### elDuderino81

Hmm, ok. So kinetic energy is conserved, so:

V^2=U^2+2as â†’ 0+2*10ms'1*50m=1000ms-1

Velocity will be sqroot = 31.62ms-1?

4. Dec 23, 2013

### SteamKing

Staff Emeritus
If the slope is frictionless, what reaction force could there be which would retard the motion of the cart?

5. Dec 23, 2013

### elDuderino81

wouldn't it normally be the perpendicular component of MgXsinÎ¸? This is what confused me initially. If it was a standard freebody diagram on a horizontal plain, the reaction force would be equal to Mass*Gravity (R=mg). .

The total energy is conserved as a gain in kinetic energy through a loss in potential. This is where I was able to make sense of using the V^2=U^2+2as, but because their isn't an initial velocity, it would simplify to the root of 2*g*Î”h.

Have I understood this properly? My answer appeared quite sensible (31.62ms-1)

6. Dec 23, 2013

### Staff: Mentor

Total mechanical energy is conserved.

That's a correct equation, but I'm not sure your reasoning is correct. That's usually a kinematic equation.

What you want is energy conservation:
KEi + PEi = KEf + PEf
PEi = KEf

mgh = 1/2mv^2

(That's equivalent to what you have done, of course.)

7. Dec 23, 2013

### Staff: Mentor

There will certainly be a normal force acting on the coaster at every point along the path. But that force is perpendicular to the velocity, so it does no work and does not affect the speed of the coaster.

Good.

8. Dec 23, 2013

### elDuderino81

Thank you Doc Al and SteamKing for prompting me in the right direction, very much appreciated! :-)

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