krypticlemonjuice
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How do we do this q?
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Mod 5 q (1 Viewer)
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How do we do this q?
This question can effectively be broken into two 'pieces'
Because the net acceleration is directly downwards at the peak of the curvature, the rollercoaster is effectively in uniform circular motion at this instant. Hence, we can use the formula for uniform circular motion:
The acceleration is gravity (9.8), and the radius is 8 (given), so using the formula we can determine the velocity to be 8.85.
Because the rollercoaster has negligible speed initially and no friction, all of the kinetic energy comes from the change in height. This means that we just need to find the height from any fall ending in the same velocity, irrespective of the horizontal movement.
The initial velocity is zero, acceleration is 9.8, we don't yet know t and we need to find s.
And hence we can use the equation from earlier
So the answer is A
Because the net acceleration is directly downwards at the peak of the curvature, the rollercoaster is effectively in uniform circular motion at this instant. Hence, we can use the formula for uniform circular motion:
The acceleration is gravity (9.8), and the radius is 8 (given), so using the formula we can determine the velocity to be 8.85.
Because the rollercoaster has negligible speed initially and no friction, all of the kinetic energy comes from the change in height. This means that we just need to find the height from any fall ending in the same velocity, irrespective of the horizontal movement.
The initial velocity is zero, acceleration is 9.8, we don't yet know t and we need to find s.
And hence we can use the equation from earlier
So the answer is A
howcanibesmarter
Well-Known Member
Another method will be using gravitational potential and centripetal force.
We know that at the top of the loop, mv^2/r = mg, solving for v gives you v=sqrt(rg)
We also know that the speed at that point of the ramp will be KE=U, 1/2mv^2final = mg change in h since Initial V =0 , solving for v gives sqrt(2gh)
Equating the two gives sqrt(rg)=sqrt(2gh), then rg=2gh, cancelling g gives r=2h.
Therefore h is r/2 = 8/2 = 4m. Therefore answer is A.
We know that at the top of the loop, mv^2/r = mg, solving for v gives you v=sqrt(rg)
We also know that the speed at that point of the ramp will be KE=U, 1/2mv^2final = mg change in h since Initial V =0 , solving for v gives sqrt(2gh)
Equating the two gives sqrt(rg)=sqrt(2gh), then rg=2gh, cancelling g gives r=2h.
Therefore h is r/2 = 8/2 = 4m. Therefore answer is A.