I found the "easy question" simple but really struggled with q19, how did u guys go

Sorry mods if wrong spot

Q19. Most part of the question 19 is ok. (total 18 marks)

Plotting the graph is straight forward (just to time mass in kg x 10 to convert to force, The compression in mm in x axis.

When you plot the graph you'll get two linear sections. Gradient of first linear graph is the K of spring A (which is 150 N/m). The gradient of the second liner section is the sum of K values of A and B which was 500 N/m. Hence the K of the spring B = 350 N/m. The scales need to be selected to spread your data more than 50% of the area. Uncertainty in compression was +/- 2.0mm . If you have selected scales appropriately, the error bars are like half a box left and right.

(c) (i) Area under the graph of spring A = 1/2*150*0.08^2 = 0.48J

C(ii) Spring Pot energy stored in the spring when the system is compressed by 80 mm. Need to calculate the area under both straight lines= 1/2*0.06*9 + (1/2(9+19)*0.02 =0.55 J

(iii) Work done only by spring B to compress to 80 mm = 1/2 * 350*0.02^2 (because Spring B only compresses from 60 mm to 80 mm). = 0.07 J

(b) How these type spring system can be used in small bumps and severe bumps.

Low K spring for small bumps and combination of Both springs for sever bumps to absorbs the shock.