Why is not A? (1 Viewer)

[kkk579]

Since ke =1/2mv^2, ke is largest with v is larger. And v is larger when acceleration is larger. Acceleration is larger when force is larger. At 4seconds it experiences the greatest net force so why is A considered wrong?

 

[Sylfiphy]

Since ke =1/2mv^2, ke is largest with v is larger. And v is larger when acceleration is larger. Acceleration is larger when force is larger. At 4seconds it experiences the greatest net force so why is A considered wrong?View attachment 43188

omg I remember this paper it was horrific

 

[liamkk112]

Since ke =1/2mv^2, ke is largest with v is larger. And v is larger when acceleration is larger. Acceleration is larger when force is larger. At 4seconds it experiences the greatest net force so why is A considered wrong?View attachment 43188

its not at 4s, it's at 7s. the mass will keep accelerating as there's always a force present in a singular direction

 

[liamkk112]

its not at 4s, it's at 7s. the mass will keep accelerating as there's always a force present in a singular direction

think of it like a simplified car (ignore friction and all that resistive stuff). you press on the accelerator at a certain magnitude , it speeds up, and then for a short while the acclerator is still at that magnitude and your car experiences a constant acceleration (4s-5s on the graph). then you let go of the accelerator, the acceleration slows down to 0, but there's no de accleration so the car keeps accelerating until the acceleration is 0. notice there was never any negative acceelration, so at the end of the graph the velocity is maximum.

 

[liamkk112]

are any of these options even correct

 

[Realstudytips]

are any of these options even correct

I think B is correct but OP crossed it out so maybe I’m buggin but I thought u js do the area under the curve

 

[wizzkids]

There is no correct answer. All the options are false.
The best way to analyse this force-time graph is to recognise that the area under the force-time graph is equal to the Impulse, which is also equal to the Change of Momentum.
Break the graph into three regular shapes, a triangle, a rectangle and another triangle.
The sum of the areas comes to 10+5+5 = 20 N.s which is also equal to mv_f, the mass multiplied by the final velocity.
So the final velocity is 4 m/s

 

[kkk579]

Yeah B is the correct one i just accidentally crossed it out first then put circled it again @Realstudytips @liamkk112

 

[kkk579]

omg I remember this paper it was horrific

HAHA from peak 2021 prelims paper the teacher told us the average for this paper was like 20-30%

 

[kkk579]

think of it like a simplified car (ignore friction and all that resistive stuff). you press on the accelerator at a certain magnitude , it speeds up, and then for a short while the acclerator is still at that magnitude and your car experiences a constant acceleration (4s-5s on the graph). then you let go of the accelerator, the acceleration slows down to 0, but there's no de accleration so the car keeps accelerating until the acceleration is 0. notice there was never any negative acceelration, so at the end of the graph the velocity is maximum.

Ohhhh yes i get it thank you but after reading @wizzkids comment none of these are even correct

Also maximum acc is between 4 and 5 right

 

[kkk579]

Also for this q shouldnt it be like this since thats the rope A is actually attached to?

 

[Realstudytips]

Also for this q shouldnt it be like this since thats the rope A is actually attached to?View attachment 43198View attachment 43197

A has two weights attached on its rope which means there are tension in two directions applied on each of the rope . I’m assuming ur free body diagram is the blue which only shows the normal and weight forces acting on it

 

[kkk579]

A has two weights attached on its rope which means there are tension in two directions applied on each of the rope . I’m assuming ur free body diagram is the blue which only shows the normal and weight forces acting on it

But the given sol doesnt show the tension which acts on the rope which is directly attached to the mass

 

[wizzkids]

Question 15 (a) is also badly worded.
I think what the examiner meant to say was, "Draw the force-vector diagram at the point Y."
If you answer the question literally, the free-body diagram for the mass A is trivial, as @kkk579 has shown, although quantifying the forces is not quite so trivial.
It is just two vectors, the upwards tension A-Y and the downwards weight force m_Ag, which are equal and opposite.
For those who are interested in the equilibrium solution to the forces at point Y, here is the answer.

 

[kkk579]

Where did i go wrong with c? The answer is 12.5



Shouldnt the forces be negative since it says the force sensors measure the force applied onto the car during collisions? So it would be towards the left direction assuming the car was moving in the right direction towards the wall.

 

[wizzkids]

@kkk579 In question 6 (c) you performed a calculation based on energy. The correct way to calculate the height is to use v² =u² +2as
Make v = 15.65 m/s, let u =0 and rearranging terms we can say S = 15.65² / 2 x 9.8 and the answer is 12.49 m

 

[kkk579]

@kkk579 In question 6 (c) you performed a calculation based on energy. The correct way to calculate the height is to use v² =u² +2as
Make v = 15.6 m/s, let u =0 and rearranging terms we can say S = 15.6² / 2 x 9.8 and the answer is 12.3 m

If u were to use energy how would u correctly do it?

 

[wizzkids]

Yes, you could do the calculation that way. If the object has half it's final velocity then it has a quarter of its final kinetic energy, and therefore it has lost a quarter of its gravitational potential energy, so it has fallen 1/4 x 50 metres, and it is 37.5 metres above the ground.

 

[kkk579]

Yes, you could do the calculation that way. If the object has half it's final velocity then it has a quarter of its final kinetic energy, and therefore it has lost a quarter of its gravitational potential energy, so it has fallen 1/4 x 50 metres

But whats wrong with my w/o?

 

[wizzkids]

@kkk579 The answer given in the paper is wrong, btw.
There is nothing fundamentally wrong with your calculation, but there is no need to include the mass of the block.
All objects fall at the same rate regardless of their mass.
You can use the standard equations for uniformly accelerated motion.