DP Physics (last assessment 2024)

Question 19M.2.SL.TZ2.biii

Date	May 2019	Marks available	[Maximum mark: 2]	Reference code	19M.2.SL.TZ2.biii
Level	SL	Paper	2	Time zone	TZ2
Command term	Determine	Question number	biii	Adapted from	N/A

biii.

[Maximum mark: 2]

19M.2.SL.TZ2.biii

A student strikes a tennis ball that is initially at rest so that it leaves the racquet at a speed of 64 m s^–1. The ball has a mass of 0.058 kg and the contact between the ball and the racquet lasts for 25 ms.

The student strikes the tennis ball at point P. The tennis ball is initially directed at an angle of 7.00° to the horizontal.

The following data are available.

Height of P = 2.80 m

Distance of student from net = 11.9 m

Height of net = 0.910 m

Initial speed of tennis ball = 64 m s^-1

Determine the speed of the tennis ball as it strikes the ground.

[2]

Markscheme

ALTERNATIVE 1

Initial KE + PE = final KE /

$\frac{1}{2}$ × 0.058 × 64² + 0.058 × 9.81 × 2.80 = $\frac{1}{2}$ × 0.058 × v² ✔

v = 64.4 «ms⁻¹» ✔

ALTERNATIVE 2

${v_v} = « \sqrt {{{7.8}^2} + 2 \times 9.81 \times 2.8} » = 10.8 « {\text{m}} {{\text{s}}^{ - 1}}$ » ✔

« $v = \sqrt {{{63.5}^2} + {{10.8}^2}}$ »

$v = 64.4 « {\text{m}} {{\text{s}}^{ - 1}}$ » ✔

Examiners report

This proved difficult for candidates at both HL and SL. Many managed to calculate the final vertical component of the velocity of the ball.

Syllabus sections

Core

Core » Topic 2: Mechanics » 2.3 – Work, energy, and power

Core » Topic 2: Mechanics

Question 19M.2.SL.TZ2.biii

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