June 2018 and 2019 Paper 1 GCE A/L Pure Mathematics with Mechanics - Introduction to GCE A/L 2019 Solution
June 2018 and 2019 Paper 1 GCE A/L Pure Mathematics with Mechanics: Introduction Topics
Introduction
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Pure Math Mech June 2019 Paper 1 Solutions
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Pure Math Mech June 2018 Paper 1 Solutions
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Page 3
Page 2
Solution to June 2018 qn 7 by Johan
This is thé solution to the question (# Q6 PMM GCE jeune 2018)
Question 7 June 2018
Using : P/v = 14.28×1000/6m/s
=2371.6N
a- considering the car and the carriage as a single body the acceleration of the system
F - R = ma
2371.6 - (639 - 280) = ( 1400 +700)a
2021.6 = 2100a
a= 0.96m/s^2
b- T - 280 = 700a
T - 280 = 490
T = 770N
c- F = P/V
= 14.28 × 1000/12
= 1190N
So, F - R = ma
1190-(630-230) = 1400+ 700a
280 = 2100a
a = 0.13 m/s^2
d- time taken ,
We have first,280=700a
a=-0.4m/s^2
Retardation of its carriage at rest
Using
V= u + at
0= 12 + ( -0.4)t
-12= -0.4t
T= 30 s
Faith's solution
That was good Faith but you made a mistake when calculating the acceleration in the (a) part of the question. You were supposed to add the resistances not subtract them. F-R=ma, R = 639 + 280 = 919, a = (F-R)/m = (2371.6-919)/(1400+700) = 0.7 in meters per squared seconds
This is the solution to the question on impacts (# GCE june 2018 Q7)
You've not uploaded any solution Rene, please do so this evening
Here's the continuation.
Nice try Winnie
Your solutions are all correct Winnie. That was good
Good evening. Here's my solution to Maths mechanics June 2018 Q7
Your solutions are all correct Winnie. But your diagram has a wrong force direction, the tension on the carried should be pointing to the right not to the left.
Your solutions are all correct Winnie. But your diagram has a wrong force direction, the tension on the carriage should be pointing to the right not to the left.