A-Level Physics Problem - Velocity

[elDuderino81]

1. A Level Physics - finding resultant velocity at bottom of a frictionless incline, with no angle provided?
2. Hi, I am currently working through a problem set and I'm being asked to find the speed of a cart (roller-coaster) after it has traveled 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 height 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-13. Am I being a bit dim here? The way I see it, without the angle of inclination I won't 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.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Doc Al]

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

Hint: What's conserved?

 

[elDuderino81]

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

V^2=U^2+2as → 0+2*10ms'1*50m=1000ms-1

Velocity will be sqroot = 31.62ms-1?

 

[SteamKing]

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

 

[elDuderino81]

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

wouldn't it normally be the perpendicular component of MgXsinθ? 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*Δh.

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

 

[Doc Al]

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

Total mechanical energy is conserved.

V^2=U^2+2as → 0+2*10ms'1*50m=1000ms-1

Velocity will be sqroot = 31.62ms-1?

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.)

 

[Doc Al]

wouldn't it normally be the perpendicular component of MgXsinθ? 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). .

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.

The total energy is conserved as a gain in kinetic energy through a loss in potential.

Good.

 

[elDuderino81]

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