2022 H2 Physics Paper 2 Q2 — Up and Down a Rough Slope, and a Collision From a Graph

The model, live

Weight componentFrictionP / object XObject Y

Part (a)(i)

1 mark

State what is meant by a vector quantity.

A quantity that has both magnitude and direction.

Part (a)(ii)

1 mark

Name one vector quantity.

Any one of: displacement, velocity, acceleration, force, momentum.

Part (b)(i)

1 mark

An object of weight $16\,\text{N}$ slides down a rough $35^\circ$ slope at constant speed. Find the component $W$ of the weight down the slope.

W = 16\sin 35^\circ = 9.18\,\text{N}

$W = 9.2\,\text{N}$

Watch

The component along the slope uses $\sin$ of the slope angle; the component into the slope uses $\cos$ . A quick test: on a flat surface ( $0^\circ$ ) nothing acts along the slope, and $\sin 0^\circ = 0$ .

Part (b)(ii)

2 marks

The object is pulled up the slope at constant speed by a force $P$ parallel to the slope. Find $P$ .

Approach

Constant speed both times means zero resultant force. Going down, the friction balances the weight component, so it is $9.2\,\text{N}$ up the slope. The normal force has not changed, so the friction is still $9.2\,\text{N}$ ; but friction opposes the motion, so going up it acts down the slope, on the same side as the weight component.

P = W + F = 9.18 + 9.18 = 18.4\,\text{N}

$P = 18\,\text{N}$

Part (c)(i)

2 marks

X ( $0.22\,\text{kg}$ ) and Y ( $0.40\,\text{kg}$ ) collide along a line. From the velocity–time graph of X, find the impulse on X.

Approach

Impulse is the change in momentum. Read X's velocity before ( $2.6\,\text{m s}^{-1}$ ) and after ( $3.5\,\text{m s}^{-1}$ ) the steep section.

\text{impulse} = m\Delta v = (0.22)(3.5 - 2.6) = 0.198\,\text{N s}

impulse $= 0.20\,\text{N s}$

Part (c)(ii)

2 marks

Before the collision Y moved at $3.3\,\text{m s}^{-1}$ . Find the speed $v$ of Y after the collision.

Approach

X sped up, so Y pushed it forwards: Y was behind X, moving the same way and faster. By Newton's third law Y receives an equal and opposite impulse, $0.198\,\text{N s}$ backwards.

v = 3.3 - \frac{0.198}{0.40} = 2.8\,\text{m s}^{-1}

To check your answer:

Total momentum before: $(0.22)(2.6) + (0.40)(3.3) = 1.89\,\text{N s}$ . After: $(0.22)(3.5) + (0.40)(2.8) = 1.89\,\text{N s}$ . Conserved, and Y ends up slower than X, so they separate.

$v = 2.8\,\text{m s}^{-1}$

Part (c)(iii)

2 marks

Find the distance travelled by X while it is in contact with Y.

Approach

Contact lasts while X's velocity is changing: from $1.8\,\text{s}$ to $2.0\,\text{s}$ . The distance is the area under the graph over that interval, a trapezium.

s = \tfrac12(2.6 + 3.5)(2.0 - 1.8) = 0.61\,\text{m}

distance $= 0.61\,\text{m}$

What this question was really testing

Resolving on a slope. $mg\sin\theta$ along, $mg\cos\theta$ into it.
Friction opposes motion. Same size if the normal force is unchanged, opposite direction when the motion reverses.
Impulse = change in momentum, equal and opposite on the two bodies.
Area under v–t is displacement, even over a fraction of a second.

Friction that changes sides, and a collision read off a velocity–time graph

Part (a)(i)

Part (a)(ii)

Part (b)(i)

Part (b)(ii)

Part (c)(i)

Part (c)(ii)

Part (c)(iii)

What this question was really testing

This is how we mark, and how we teach.