Unit 4 Physics Edexcel A2 and Edexcel IAL

A car is repeatedly driven over a bridge at gradually increasing speeds. Above a certain speed the car loses contact with the road at X. State why this happens.

The mark scheme says: An answer that either states implicitly or implies that 'The required centripetal force > mg and so cannot be provided'.

but i didn't really understand what they meant

Reply 261

mathphymechster

Original post by target21859

Hi, I have two questions. For the first picture how do you know which way the current is flowing? And for the second picture why is the answer B? Thanks in advance.

First pic: The diagram indicates that the track is an antiproton's, and from content knowledge you should be able to deduce that antiprotons (antiparticle of proton) have a negative charge. Hence even though the (-) particles travel forwards, the (conventional +) current flows backwards (to the left of the page).

Second pic: Faraday's Law: ε= dφ/dt. The magnitude of the e.m.f is given by the rate of change of magnetic flux against time. Simpler put, find the gradient of the curve of [magnetic flux against time]. Overall the curve gets steeper upwards, hence the e.m.f. increases with time. This eliminates all but A and B, and between the two you look at the curve at t=0 to verify whether it starts completely flat (like a parabola, gradient=0) or starts at a gradient (>0) already. In this case, it's the latter, which corresponds to graph B.

Reply 262

kritz20

2

What is the minimum requirement for an A* in physics?

Reply 263

swinroy

12

Original post by physicsmaths

Coil X has a field which is also acting on Coil Y with the same magnirude. So it would be 0.002x50x4.

Posted from TSR Mobile

BAN = Flux Linkage = 0.002 x 0.0004 x 50
= 0 0.00004 (answer C)
have to convert sq cm to sq m correctly

Reply 264

Shoot

7

Original post by dnaamani

A car is repeatedly driven over a bridge at gradually increasing speeds. Above a certain speed the car loses contact with the road at X. State why this happens.

The mark scheme says: An answer that either states implicitly or implies that 'The required centripetal force > mg and so cannot be provided'.

but i didn't really understand what they meant

The centripetal force is basically the force that acts perpendicular to the direction of a body and that is directed inward toward the centre of rotation. It's important to remember that the centripetal force isn't an independent force either. In this scenario the weight (mg) is the centripetal force.

If we look at the equation, Fc=mv^2/r, increasing V, means that the centripetal force required to keep the car in contact with the road increases. AS the magnitude of the centripetal force required > mg , the car loses contact with the road as mg < than the required centripetal force.

Does that make sense? Or would you like further clarification.

Reply 265

swinroy

12

Answer to Q8 second picture is B because: Faraday's law of electromagnetic induction (EMF = Rate of Change of flux linkage)
The flux is changing like a quadratic function of time (parabolic) so answer is either A or B since rate of change of quadratic is a linear graph.
The EMF is rising linearly because the gradient of flux graph is rising linearly.
If flux graph started on a minimum, (ie FLAT) answer would be A
But there is a non-zero initial flux gradient so answer is B

first bit
ANTIPROTON = the antiparticle of the proton; a particle having the same mass as the proton but an equal and opposite charge
so when a NEGATIVE charge travels from left to right, the conventional current direction used in Fleming's hand rules is right to left

Reply 266

YAH_BOY_ASH

7

Original post by k92e67

Im wondering too. Anyone know?

Average of 80% UMS in all units and 90% UMS average in A2 units

Reply 267

Believerrrr

9

If a positively charged particle enters magnetic field ,current is in the direction of the moving direction of the particle .And if an electron (negatively charged particle) enters the magnetic field ,the current always flows in the opposite direction .

Reply 268

Believerrrr

9

Original post by Shoot

The centripetal force is basically the force that acts perpendicular to the direction of a body and that is directed inward toward the centre of rotation. It's important to remember that the centripetal force isn't an independent force either. In this scenario the weight (mg) is the centripetal force.

If we look at the equation, Fc=mv^2/r, increasing V, means that the centripetal force required to keep the car in contact with the road increases. AS the magnitude of the centripetal force required > mg , the car loses contact with the road as mg < than the required centripetal force.

Does that make sense? Or would you like further clarification.

Hey ! I did the same question today and had the same doubt .i get the point that resultant force increases as the centripetal force increases .but to which direction ? Centripetal force always acts towards the centre right ? So how could the car get a force upwards to lose contact with the ground ? :s-smilie:

Reply 269

ElvishJumpSuit

7

Original post by Believerrrr

Hey ! I did the same question today and had the same doubt .i get the point that resultant force increases as the centripetal force increases .but to which direction ? Centripetal force always acts towards the centre right ? So how could the car get a force upwards to lose contact with the ground ? :s-smilie:

The forces acting (in the middle) are the weight of the car downward and normal reaction up.

The resultant of these real forces (towards the centre) must equal mv²/r (Newton's II Law) F=ma

mg + (-R) = mv²/r (-R) because it points away from centre

so R = mg - mv²/r

R represents normal contact force with the road, and it will fall to zero as the car is about to leave the road (when mg = mv²/r)

The geometry of the bridge and the speed of the car create a minimum requirement of a force towards the centre of the curve. Only at the top of the curve is weight pointing towards the centre (making the above equations valid as the forces line up). At this point, there is either enough force or there isn't. If weight is not enough to make the car follow the curve, then it won't. There is no need for an upward force to make it leave the road. The road just has to curve away from under it. Just like there doesn't need to be an extra force to make a car skid going round a corner, just not enough of a particular one (Friction).

Hope this helps!

My students used to do this course and my YouTube channel has lots of video walkthroughs of past papers from Unit 4 EDEXCEL.
I'm pretty sure I have done this one before.

(Just don't ask me to do any new ones!!!)

EJS

Reply 270

Believerrrr

9

Original post by ElvishJumpSuit

The forces acting (in the middle) are the weight of the car downward and normal reaction up.

The resultant of these real forces (towards the centre) must equal mv²/r (Newton's II Law) F=ma

so R = mg - mv²/r

R represents normal contact force with the road, and it will fall to zero as the car is about to leave the road (when mg = mv²/r)

The geometry of the bridge and the speed of the car create a minimum requirement of a force towards the centre of the curve. Only at the top of the curve is weight pointing towards the centre (making the above equations valid as the forces line up). At this point, there is either enough force or there isn't. If weight is not enough to make the car follow the curve, then it won't. There is no need for an upward force to make it leave the road. The road just has to curve away from under it. Just like there doesn't need to be an extra force to make a car skid going round a corner, just not enough of a particular one (Friction).

Hope this helps!

My students used to do this course and my YouTube channel has lots of video walkthroughs of past papers from Unit 4 EDEXCEL.
I'm pretty sure I have done this one before.

(Just don't ask me to do any new ones!!!)

EJS

Got it .Thank you so much :smile: