omg I remember this paper it was horrificSince 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 directionSince 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
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.its not at 4s, it's at 7s. the mass will keep accelerating as there's always a force present in a singular direction
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 curveare any of these options even correct
HAHA from peak 2021 prelims paper the teacher told us the average for this paper was like 20-30%omg I remember this paper it was horrific
Ohhhh yes i get it thank you but after reading @wizzkids comment none of these are even correctthink 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.
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 itAlso for this q shouldnt it be like this since thats the rope A is actually attached to?View attachment 43198View attachment 43197
But the given sol doesnt show the tension which acts on the rope which is directly attached to the massA 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
If u were to use energy how would u correctly do it?@kkk579 In question 6 (c) you performed a calculation based on energy. The correct way to calculate the height is to use v^{2} =u^{2} +2as
Make v = 15.6 m/s, let u =0 and rearranging terms we can say S = 15.6^{2} / 2 x 9.8 and the answer is 12.3 m
But whats wrong with my w/o?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