Mathematics 9709/43 — May/June 2022

Cambridge AS Level · Mechanics · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Kinematics of Motion in a Straight Line · Forces and Equilibrium · Newton's Laws of Motion · Momentum · Energy, Work and Power

Q1MomentumFree sample

Two particles $P$ and $Q$ , of masses $0.3\text{ kg}$ and $0.2\text{ kg}$ respectively, are at rest on a smooth horizontal plane. $P$ is projected at a speed of $4\text{ m s}^{-1}$ directly towards $Q$ . After $P$ and $Q$ collide, $Q$ begins to move with a speed of $3\text{ m s}^{-1}$ .

(a)

Find the speed of $P$ after the collision.

DifficultyMedium-Easy

Worked solution

Approach

Use conservation of linear momentum in the direction of motion. Since the plane is smooth, no external horizontal force acts during the collision, so total momentum before equals total momentum after. Let $v$ be the speed of $P$ after the collision, and take the direction of $P$ 's initial motion as positive.

Working

Before collision:

0.3 \times 4 + 0.2 \times 0 = 1.2

After collision:

0.3v + 0.2 \times 3 = 0.3v + 0.6

Equating:

1.2 = 0.3v + 0.6

0.3v = 0.6

v = 2

Answer

The speed of $P$ after the collision is $2\text{ m s}^{-1}$ .

Final answer

2 m s^{-1}

Detailed explanation

Walkthrough

We are told $P$ and $Q$ are on a smooth horizontal plane, so there is no friction or other horizontal external force. During the collision the only significant forces are internal between $P$ and $Q$ , so total momentum is conserved. Choose a positive direction, the direction in which $P$ was projected. Write momentum as mass multiplied by velocity. Before the collision only $P$ moves: $0.3 \times 4$ . After the collision both particles move: $P$ at unknown speed $v$ and $Q$ at $3\text{ m s}^{-1}$ . Set momentum before equal to momentum after and solve for $v$ . The positive result confirms $P$ continues in the same direction.

Key Takeaways

Momentum is conserved in a direct collision when no external horizontal force acts. Momentum is a vector, so a positive direction must be chosen. On a smooth plane the surface exerts no horizontal force, so horizontal momentum is conserved during the collision.

Common Mistakes

Forgetting that $Q$ is initially at rest and has zero momentum. Making sign errors by not assigning a positive direction. Not writing the conservation equation, which is required for the method mark. Making an arithmetic error when solving $1.2 = 0.3v + 0.6$ .

Things to Be Careful About

Use consistent units of kg and m/s. The smooth plane supports the weights vertically, but vertical forces balance, so only horizontal momentum is conserved. If $v$ had been negative it would mean $P$ reverses direction; here it is positive.

Techniques used

apply conservation of linear momentumsolve linear equation for unknown speed

(b)

After the collision, $Q$ moves directly towards a third particle $R$ , of mass $m\text{ kg}$ , which is at rest on the plane. The two particles $Q$ and $R$ coalesce on impact and move with a speed of $2\text{ m s}^{-1}$ .

Find $m$ .

DifficultyMedium-Easy

Worked solution

Approach

After the first collision, $Q$ moves at $3\text{ m s}^{-1}$ towards stationary $R$ . When $Q$ and $R$ coalesce, they move together as a single particle of mass $0.2 + m$ at $2\text{ m s}^{-1}$ . Use conservation of linear momentum in the same direction.

Working

Before $Q$ and $R$ collide:

0.2 \times 3 + m \times 0 = 0.6

After coalescence:

(0.2 + m) \times 2

Equating:

0.6 = (0.2 + m) \times 2

0.6 = 0.4 + 2m

2m = 0.2

m = 0.1

Answer

m = 0.1\text{ kg}

Final answer

m = 0.1 kg

Detailed explanation

Walkthrough

After part (a), $Q$ has speed $3\text{ m s}^{-1}$ . $R$ is at rest, so its initial momentum is zero. $Q$ and $R$ coalesce, so after impact they form one body of mass $0.2 + m$ moving at $2\text{ m s}^{-1}$ . Since the collision is again on a smooth horizontal plane, momentum is conserved horizontally. Set momentum before equal to momentum after and solve for $m$ . This is a direct application of conservation of momentum to a coalescing collision.

Key Takeaways

When particles coalesce, add their masses and use the common final speed. Momentum conservation applies to all direct impacts, including coalescence. A stationary particle contributes zero momentum.

Common Mistakes

Using only $Q$ 's mass after coalescence instead of $0.2 + m$ . Forgetting that $R$ is initially at rest. Not multiplying the combined mass by the final speed. Making an algebra error when solving $0.6 = (0.2 + m) \times 2$ .

Things to Be Careful About

Keep units in kg and m/s. The final speed $2\text{ m s}^{-1}$ is for the combined particle, so the mass must be the sum. The result $m = 0.1\text{ kg}$ is positive and sensible. The mark scheme requires the conservation equation to be written before solving.

Techniques used

apply conservation of linear momentum to coalescing particlessolve for unknown mass

The rest of this paper

6 more questions

Q2Kinematics of Motion in a Straight Line5M
Q3Kinematics of Motion in a Straight Line5M
Q4Forces and Equilibrium8M
Q5Newton's Laws of Motion · Energy, Work and Power8M
Q6Newton's Laws of Motion · Forces and Equilibrium10M
Q7Kinematics of Motion in a Straight Line10M

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