Recent content by Mr.A.Gibson

Thanks, that's the part I missed.

Surely that will give you acceleration and velocity as a function of displacement not time? Since F and x vary with time i can't see how to make the differential work, in fact I cannot get any equation as a function of time.

I'm ok with this question up until a3, I have no idea how to get velocity as a function of time from the information available. This question is taken from the Oxford Physics entrance exam. I'm not sure if it's a miss-print, perhaps the questions should be as a function of x, because that seems...

When at terminal velocity the forces are in balance Bouancy = weight + drag Put in the formula you have and then rearrange for v. Then show an attempted solution.

You calculate the drag force from the eqation, it's components are: ρ the density of air, v the velocity of the balloon, when the forces are balanced this is the terminal velocity A cross-sectional area of the balloon, calculate from the radius C_d Drag coefficient, this changes for different...

But it will not accelerate at this rate for any appreciable length of time. When it starts to move you have to consider air resistance. Using the drag equation the terminal velocity of your balloon is 7.3m/s, so it will not get any faster than this.

2ħk = 10m m = ħk/10 Makes sense, what did I mess up in the first place then?

conservation of momentum, p_b+10m=p_a subscripts refers to photons momentum before and after collision. Then from conservation of energy \frac{1}{2}mv^2+cp_b=cp_a substitute in the first equation and velocity of the object to give 50m+cp_b=c(p_b+10m) 50m=10cm only valid if m=0

I can find no fault in your calculation. Of course the moment it started to move, air-resistance would come into effect and the balloon would very quickly reach terminal velocity.

Homework Statement The momentum of a photon is ħk, where k is the wave vector. Assume an object moving at 10m/s, objects of which mass can be stopped by elastic reflection of the photon. Homework Equations Conservation of energy and momentum p = ħk = h/λ The Attempt at a Solution...

We should really consider his weight will be acting at this point. Hence the free body diagram will have the force of his weight, and the tension in the two ropes acting on his hands. The components of the tension you seem to have done correctly. Part C, moment = force x perpendicular distance...

Your method is correct as far as I can tell. Two photons would have to be produced to conserve momentum but the question does ask for the total energy. Perhaps the number of marks is a misprint.

Good question, the energy is stored is due to the tension on a molecular level. As the energy move down the rope it moves from particle to particle. If we imagine the rope as a line of single particles. Then as one particle moves the bond to the next particle is stretched like a spring...

You must also consider the energy stored in the spring E = \frac{1}{2}kx^2 The total potential energy lost by the box during the fall, is equivalent to the energy stored in the spring at the moment when the box is at rest. This will allow you to calculate k. Then use the total energy...

At the top of the loop the forces acting on the pilot are the seat pushing on him and gravity. Hence the total force is F = mg + 0.25mg = 1.25mg also F = ma = mv^2/r equate these two and resolve for v At the bottom of the loop, again calculate the resultant force and use the same method...