DP Physics (last assessment 2024)

Question 21M.2.SL.TZ1.1

Date	May 2021	Marks available	[Maximum mark: 11]	Reference code	21M.2.SL.TZ1.1
Level	SL	Paper	2	Time zone	TZ1
Command term	Calculate, Determine, Show that, Sketch, State	Question number	1	Adapted from	N/A

[Maximum mark: 11]

21M.2.SL.TZ1.1

Two players are playing table tennis. Player A hits the ball at a height of 0.24 m above the edge of the table, measured from the top of the table to the bottom of the ball. The initial speed of the ball is 12.0 m s⁻¹ horizontally. Assume that air resistance is negligible.

(a)

Show that the time taken for the ball to reach the surface of the table is about 0.2 s.

[1]

Markscheme

t = « $\sqrt{\frac{2 d}{g}}$ =» 0.22 «s»
OR

t = $\sqrt{\frac{2 \times 0.24}{9.8}}$ ✓

Answer to 2 or more significant figures or formula with variables replaced by correct values.

(b)

Sketch, on the axes, a graph showing the variation with time of the vertical component of velocity v_v of the ball until it reaches the table surface. Take g to be +10 m s⁻².

[2]

Markscheme

increasing straight line from zero up to 0.2 s in x-axis ✓

with gradient = 10 ✓

(c)

The net is stretched across the middle of the table. The table has a length of 2.74 m and the net has a height of 15.0 cm.

Show that the ball will go over the net.

[3]

Markscheme

ALTERNATIVE 1

$t = \frac{1.37}{12} =$ «0.114 s» ✓

$y = \frac{1}{2} \times 10 \times 0 . 114^{2} = 0.065$ m ✓

so (0.24 − 0.065) = 0.175 > 0.15 OR 0.065 < (0.24 − 0.15) «so it goes over the net» ✓

ALTERNATIVE 2

«0.24 − 0.15 = 0.09 = $\frac{1}{2} \times 10 \times t^{2}$ so» t = 0.134 s ✓

0.134 × 12 = 1.6 m ✓

1.6 > 1.37 «so ball passed the net already» ✓

Allow use of g = 9.8.

The ball bounces and then reaches a peak height of 0.18 m above the table with a horizontal speed of 10.5 m s⁻¹. The mass of the ball is 2.7 g.

(d.i)

Determine the kinetic energy of the ball immediately after the bounce.

[2]

Markscheme

ALTERNATIVE 1

KE = $\frac{1}{2}$ mv² + mgh = $\frac{1}{2}$ 0.0027 ×10.5² + 0.0027 × 9.8 × 0.18 ✓

0.15 «J» ✓

ALTERNATIVE 2

Use of v_x = 10.5 AND v_y= 1.88 to get v = « $\sqrt{10 . 5^{2} + 1 . 88^{2}}$ » = 10.67 «m s⁻¹» ✓

KE = $\frac{1}{2}$ × 0.0027 × 10.67² = 0.15 «J» ✓

(d.ii)

Player B intercepts the ball when it is at its peak height. Player B holds a paddle (racket) stationary and vertical. The ball is in contact with the paddle for 0.010 s. Assume the collision is elastic.

Calculate the average force exerted by the ball on the paddle. State your answer to an appropriate number of significant figures.

[3]

Markscheme

$Δ v = 21$ «m s⁻¹» ✓

$F = \frac{0.0027 \times 21}{0.01}$

5.67 «N» ✓

any answer to 2 significant figures «N» ✓

Syllabus sections

Core

Core » Topic 2: Mechanics » 2.1 – Motion

Core » Topic 2: Mechanics

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

Core » Topic 2: Mechanics » 2.4 – Momentum and impulse

Core » Topic 1: Measurements and uncertainties » 1.1 – Measurements in physics

Core » Topic 1: Measurements and uncertainties

Question 21M.2.SL.TZ1.1

Select a Test

Syllabus sections

Confirm action