DP Physics (first assessment 2025)

Question EXE.2.HL.TZ0.4

Date	Example questions Example questions	Marks available	[Maximum mark: 6]	Reference code	EXE.2.HL.TZ0.4
Level	HL	Paper	2	Time zone	TZ0
Command term	Determine, Outline, Show that, State	Question number	4	Adapted from	N/A

[Maximum mark: 6]

EXE.2.HL.TZ0.4

Two curling stones of the same mass collide elastically on a horizontal frictionless surface. Stone A moves with an initial speed v and stone B is initially stationary. The speeds of the stones after the collision are v_A and v_B. Their directions of motion make angles of 70° and 20° with the initial velocity of stone A.

(a.i)

State what is meant by an elastic collision.

[1]

Markscheme

No change in the kinetic energy of the system ✓

(a.ii)

No unbalanced external forces act on the system of the curling stones. Outline why the momentum of the system does not change during the collision.

[1]

Markscheme

From $F = \frac{Δ p}{Δ t}$ , zero net force on the system implies that $Δ p = 0$

From Newton’s third law, the impulse delivered to A is equal but opposite to the impulse delivered to B, hence $Δ p = 0$ for the system ✓

(b)

Show that $v_{B} = \frac{\sin 70 °}{\sin 20 °} v_{A}$ .

[1]

Markscheme

The vertical momentum is zero hence $m v_{A} \sin 70 ° - m v_{B} \sin 20 ° = 0$ ✓

«leading to the expected result» ✓

(c)

Determine v_A. State the answer in terms of v.

[3]

Markscheme

Energy is conserved hence $\frac{1}{2} m {v_{A}}^{2} + \frac{1}{2} m {v_{B}}^{2} = \frac{1}{2} m v^{2}$ ✓

Eliminate $v_{B}$ using the result of part (a), e.g., ${v_{A}}^{2} + {(\frac{\sin 70 °}{\sin 20 °})}^{2} {v_{A}}^{2} = v^{2}$ ✓

$v_{A} = 0.342 v$ ✓

Syllabus sections

A. Space, time and motion

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

Question EXE.2.HL.TZ0.4

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