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IB DP Physics Mock Exam SL Paper 2 Set 5 - 2025 Syllabus

IB DP Physics Mock Exam SL Paper 1B Set 5

Prepare for the IB DP Physics Mock Exam SL Paper 2 Set 5 with our comprehensive mock exam set 5. Test your knowledge and understanding of key concepts with challenging questions covering all essential topics. Identify areas for improvement and boost your confidence for the real exam

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Question

A company designs a spring system for loading ice blocks onto a truck. The ice block is placed in a holder H in front of the spring and an electric motor compresses the spring by pushing H to the left. When the spring is released the ice block is accelerated towards a
ramp ABC. When the spring is fully decompressed, the ice block loses contact with the spring at A. The mass of the ice block is 55 kg. 

                               

Assume that the surface of the ramp is frictionless and that the masses of the spring and the holder are negligible compared to the mass of the ice block.

a. (i) The block arrives at C with a speed of 0.90ms−1. Show that the elastic energy stored in the spring is 670J.

(ii) Calculate the speed of the block at A. [4]

b. Describe the motion of the block

(i) from A to B with reference to Newton’s first law.

(ii) from B to C with reference to Newton’s second law. [3]

c. On the axes, sketch a graph to show how the displacement of the block varies with time from A to C. (You do not have to put numbers on the axes.)

         [2]

d. The spring decompression takes 0.42s. Determine the average force that the spring exerts on the block. [2]
e. The electric motor is connected to a source of potential difference 120V and draws a current of 6.8A. The motor takes 1.5s to compress the spring.

Estimate the efficiency of the motor. [2] 

Answer/Explanation

Ans:

a. (i)
\(\ll {E_{{\rm{el}}}}{\rm{ = }}\gg \frac{1}{2}m{v^{\rm{2}}} + mgh\)
OR
«Eel=»EP+EK

\(\ll {E_{{\rm{el}}}}{\rm{ = }}\gg \frac{1}{2} \times {\rm{55}} \times {\rm{0.9}}{{\rm{0}}^{\rm{2}}}{\rm{ + 55}} \times {\rm{9.8}} \times {\rm{1.2}}\)

OR
669 J 
«Eel = 669 ≈ 670J».

(ii)
\(\frac{1}{2} \times {\rm{55}} \times {v^{\rm{2}}} = 670{\rm{J}}\)

\(v = \ll \sqrt {\frac{{2 \times 670}}{{55}} = } \gg 4.9{\rm{m}}{{\rm{s}}^{ – 1}}\)

If 682J used, answer is 5.0ms–1.

b.(i)

no force/friction on the block, hence constant motion/velocity/speed

b (ii)

force acts on block OR gravity/component of weight pulls down slope

velocity/speed decreases OR it is slowing down OR it decelerates

c.

straight line through origin for at least one-third of the total length of time axis covered by candidate line

followed by curve with decreasing positive gradient

d. \(F\ll  = \frac{{\Delta p}}{{\Delta t}}\gg  = \frac{{55 \times 4.9}}{{0.42}}\)

F=642≈640N

e.

«energy supplied by motor =» 120 × 6.8 × 1.5 or 1224 J
OR
«power supplied by motor =» 120 × 6.8 or 816 W
e = 0.55 or 0.547 or 55% or 54.7%

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