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Question 18M.2.HL.TZ2.e

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Date May 2018 Marks available [Maximum mark: 3] Reference code 18M.2.HL.TZ2.e
Level HL Paper 2 Time zone TZ2
Command term Determine Question number e Adapted from N/A
e.
[Maximum mark: 3]
18M.2.HL.TZ2.e

A second identical ball is placed at the bottom of the bowl and the first ball is displaced so that its height from the horizontal is equal to 8.0 m.

                                   M18/4/PHYSI/SP2/ENG/TZ2/01.d

The first ball is released and eventually strikes the second ball. The two balls remain in contact. Determine, in m, the maximum height reached by the two balls.

[3]

Markscheme

speed before collision v = « 2 g R =» 12.5 «ms–1»

«from conservation of momentum» common speed after collision is 1 2  initial speed «vc =  12.5 2 = 6.25 ms–1»

h = « v c 2 2 g = 6.25 2 2 × 9.81 » 2.0 «m»

 

Allow 12.5 from incorrect use of kinematics equations

Award [3] for a bald correct answer

Award [0] for mg(8) = 2mgh leading to h = 4 m if done in one step.

Allow ECF from MP1

Allow ECF from MP2

[3 marks]

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