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# A2 physics questions watch

1. 1) constant resulting accelerating force found earlier = 2000N
this constant resultant force acts on the moving car although it is moving through air. Sketch a graph to show how the driving force would have to vary with time to produce this constant acceleration. Explain the shape of your graph.

Question asked how the driving force varies with time, so I was a little thrown by that phrase.

2) One trolley of mass 0.25kg and another of mass 0.50kg, each of length 0.20m, are held in contact on a horizontal runwat. A spring inside one of the trolleys is released and they are forced apart with a total K.E of 0.75J
i) By considering the energy and momentum changes, calculate the velocity of each trolley immediately after separation.

I get -2m/s for trolley (mass 0.25kg) and +1m/s for trolley (mass 0.5kg)

ii) Each trolley travels a distance of 1.0m after separation, strikes a rigid vertical obstacle and on rebounding loses 19% of its kinetic energy. Ignoring both the effect of friction and the duration of the collisions with the obstacles, calculate the position on the runway at which the trolleys next collide.
I am stuck.
2. Assume a force acting on the car to be F = -bv. v = at Hence retarding force F = -bat.
Thus F = ma + bat Thus its a linear line from some non-zero F at t = 0, going upwards.

question 2 seems correct. The final total momentum = 0, which is right and KE total equals 0.75J.

Okay. The first trolly travels a for 1m/ 2ms-1 = ½s Then hits the wall and rebounds. In another ½ it covers ½v.
Thus the trolly will be 2 - ½v away from the second trolly at that time. As it takes 1 s for the second trolly to hit the wall.
v for the first trolly is
0.5J (1-0.19) = ½0.25v²
½ 0.81 = 1/8 v²
4 0.81 = v²
v = 1.8ms-1

v for the second trolly can be worked out from the same again.
0.25J (1-0.19) = ½0.5v²
0.81 = v²
v = 0.9ms-1

So the trollies approach each other at a speed of 2.7ms-1
They are (2 - 0.9)m = 1.1m apart.
Thus it takes them 1.1 / 2.7 = 0.407s for the trollies to meet.
The first trolly thus covers another 0.407 x 1.8 m = 0.73m
Thus it will have travelled 0.9m + 0.73m since bouncing. Remember the trolly is 0.2m long. The point of collision will be at (0.9 + 0.73 + 0.2)m = 1.833m from the end of the track (from the side that the first trolly is on).
3. Assume a force acting on the car to be F = -bv. v = at Hence retarding force F = -bat.
Thus F = ma + bat Thus its a linear line from some non-zero F at t = 0, going upwards.
how can the force acting on the car be in the other direction?
The constant resultant accelerating force and the driving force are in the same direction. So you have to include resistive forces as well?

question 2 seems correct. The final total momentum = 0, which is right and KE total equals 0.75J.
okay, thanks

Okay. The first trolly travels a for 1m/ 2ms-1 = ½s Then hits the wall and rebounds. In another ½ it covers ½v.
Thus the trolly will be 2 - ½v away from the second trolly at that time. As it takes 1 s for the second trolly to hit the wall.
v for the first trolly is
0.5J (1-0.19) = ½0.25v²
½ 0.81 = 1/8 v²
4 0.81 = v²
v = 1.8ms-1

v for the second trolly can be worked out from the same again.
0.25J (1-0.19) = ½0.5v²
0.81 = v²
v = 0.9ms-1

So the trollies approach each other at a speed of 2.7ms-1
They are (2 - 0.9)m = 1.1m apart.
Thus it takes them 1.1 / 2.7 = 0.407s for the trollies to meet.
The first trolly thus covers another 0.407 x 1.8 m = 0.73m
Thus it will have travelled 0.9m + 0.73m since bouncing. Remember the trolly is 0.2m long. The point of collision will be at (0.9 + 0.73 + 0.2)m = 1.833m from the end of the track (from the side that the first trolly is on).
book says 0.57m from one end.
4. (Original post by Widowmaker)
how can the force acting on the car be in the other direction?
The constant resultant accelerating force and the driving force are in the same direction. So you have to include resistive forces as well?
F = ma

therefore a constant force will give a constant acceleration

HOWEVER, the resistive force will increase as the car speeds up, hence the driving force must increase to compensate.

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