DP Physics (first assessment 2025)

Question EXE.2.HL.TZ0.3

Date	Example questions Example questions	Marks available	[Maximum mark: 10]	Reference code	EXE.2.HL.TZ0.3
Level	HL	Paper	2	Time zone	TZ0
Command term	Calculate, Determine, Hence, Predict, State	Question number	3	Adapted from	N/A

[Maximum mark: 10]

EXE.2.HL.TZ0.3

Ball A, moving in a horizontal direction at an initial speed of 2.0 m s⁻¹ collides with a stationary ball B of the same mass. After the collision, ball A moves at a speed of 1.0 m s⁻¹ at an angle of 45° to the original direction of motion.

(a.i)

State the vertical component of the total momentum of the balls after the collision.

[1]

Markscheme

Zero ✓

(a.ii)

Hence, calculate the vertical component of the velocity of ball B after the collision.

[2]

Markscheme

$m \times 1.0 \times \sin 45 ° + m \times v_{y} = 0$ ✓

$v_{y}$ = «−»0.71 «m s⁻¹» ✓

(b)

Determine the angle θ that the velocity of ball B makes with the initial direction of motion of ball A.

[3]

Markscheme

The use of conservation of momentum in the horizontal direction, e.g. $m \times 1.0 \times \cos 45 ° + m \times v_{x} = m \times 2.0$ ✓

$v_{x} =$ « $2.0 - 1.0 \times \cos 45 ° =$ » $1.3$ «m s⁻¹» ✓

$θ = \tan^{- 1} (\frac{v_{y}}{v_{x}}) = 29 °$ ✓

(c)

Predict whether the collision is elastic.

[4]

Markscheme

Initial kinetic energy $= \frac{1}{2} m {(2.0)}^{2} = 2.0 (m)$

Final kinetic energy $= \frac{1}{2} m ({(1.0)}^{2} + {(0.71)}^{2} + {(1.3)}^{2})$ ✓

$1.6 (m)$ ✓

Final energy is less than the initial energy hence inelastic ✓

Syllabus sections

A. Space, time and motion

A. Space, time and motion » A.2 Forces and momentum

Question EXE.2.HL.TZ0.3

Select a Test

Syllabus sections

Confirm action