DP Physics (first assessment 2025)

Question 19M.2.SL.TZ2.1biii

biii.

[Maximum mark: 2]

19M.2.SL.TZ2.1biii

(biii)

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

[2]

Markscheme

ALTERNATIVE 1

Initial KE + PE = final KE /

$\frac{1}{2} \times 0.058 \times {64^2} + 0.058 \times 9.81 \times 2.80 = \frac{1}{2} \times 0.058 \times {v^2}$ ✔

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

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