Mathematics 9709/41 — May/June 2023

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 · Energy, Work and Power · Newton's Laws of Motion · Momentum

Q1MomentumFree sample

Two particles $P$ and $Q$ , of masses $m\text{ kg}$ and $0.3\text{ kg}$ respectively, are at rest on a smooth horizontal plane. $P$ is projected at a speed of $5\text{ m s}^{-1}$ directly towards $Q$ . After $P$ and $Q$ collide, $P$ moves with a speed of $2\text{ m s}^{-1}$ in the same direction as it was originally moving.

(a)

Find, in terms of $m$ , the speed of $Q$ after the collision.

DifficultyMedium-Easy

Worked solution

Approach

Use conservation of linear momentum along the direction of motion. Since the plane is smooth, there is no external horizontal force during the collision, so the total momentum of $P$ and $Q$ is conserved.

Working

Take the direction of $P$ 's original motion as positive. Let $v$ be the speed of $Q$ after the collision.

m \times 5 + 0.3 \times 0 = m \times 2 + 0.3v

This simplifies to

5m = 2m + 0.3v

0.3v = 3m

v = 10m

Answer

The speed of $Q$ after the collision is $10m\text{ m s}^{-1}$ .

Final answer

10m m/s

Detailed explanation

Walkthrough

Before the collision, only $P$ is moving. Its momentum is its mass multiplied by its velocity: $m \times 5 = 5m$ . Particle $Q$ is at rest, so its momentum is zero. After the collision, $P$ is still moving in the same direction at $2\text{ m s}^{-1}$ , so its momentum is $2m$ . Let $v$ be $Q$ 's speed; then $Q$ 's momentum is $0.3v$ . Because the plane is smooth and there are no horizontal external forces, the total momentum before the collision must equal the total momentum after the collision. This gives $5m = 2m + 0.3v$ . Subtract $2m$ from both sides to get $0.3v = 3m$ , then divide by $0.3$ to get $v = 10m$ .

Key Takeaways

This question tests the conservation of linear momentum in a direct collision. Momentum is a vector, so a positive direction must be chosen and all velocities must be measured consistently in that direction. The smooth plane tells us there is no friction or external horizontal impulse, so momentum is conserved.

Common Mistakes

Writing a momentum term as weight, e.g. using $mg$ instead of $m$ times velocity. Momentum is mass $\times$ velocity, not weight.
Forgetting that $Q$ is initially at rest and contributes zero momentum.
Using inconsistent signs for velocities. Here both $P$ 's initial and final velocities are in the same direction, so both terms are positive.
Giving $v = -10m$ ; the question asks for speed, so the answer is positive.

Things to Be Careful About

Choose a clear positive direction and state it.
Keep the answer in terms of $m$ ; do not try to find a numerical value in part (a).
Include units if desired, but the algebraic expression $10m$ is the essential result.

Techniques used

apply conservation of linear momentumsolve a linear equation for an unknown speed

(b)

After this collision, $Q$ moves directly towards a third particle $R$ , of mass $0.6\text{ kg}$ , which is at rest on the plane. $Q$ is brought to rest in the collision with $R$ , and $R$ begins to move with a speed of $1.5\text{ m s}^{-1}$ .

Find the value of $m$ .

DifficultyMedium-Easy

Worked solution

Approach

Use conservation of linear momentum in the collision between $Q$ and $R$ . From part (a), $Q$ 's speed before hitting $R$ is $10m$ . Since $Q$ is brought to rest and $R$ starts from rest, the total momentum before equals the total momentum after.

Working

Take the direction of $Q$ 's motion as positive.

0.3(10m) + 0.6 \times 0 = 0.3 \times 0 + 0.6 \times 1.5

3m = 0.9

m = 0.3

Answer

$m = 0.3\text{ kg}$ .

Final answer

m = 0.3

Detailed explanation

Walkthrough

After the first collision, $Q$ has speed $10m$ . It now collides with $R$ , which is at rest. Before this second collision, only $Q$ is moving, so the total momentum is $0.3(10m) = 3m$ . After the collision, $Q$ is brought to rest, so its momentum is zero, and $R$ moves at $1.5\text{ m s}^{-1}$ , giving momentum $0.6 \times 1.5 = 0.9$ . Conservation of momentum gives $3m = 0.9$ , so $m = 0.3$ .

Key Takeaways

This part uses the result from part (a) and applies conservation of momentum again in a second direct collision. It shows that momentum conservation can be applied separately to each collision, using the correct masses and velocities at that instant.

Common Mistakes

Using the wrong masses: the collision is between $Q$ (mass $0.3$ ) and $R$ (mass $0.6$ ), not involving $P$ .
Forgetting that $Q$ is brought to rest, so its final momentum is zero.
Forgetting that $R$ is initially at rest, so its initial momentum is zero.
Using the initial speed of $P$ ( $5$ ) instead of $Q$ 's speed after the first collision ( $10m$ ).

Things to Be Careful About

If part (a) was answered incorrectly, part (b) can still be awarded follow-through marks as long as the momentum equation for $Q$ and $R$ is set up correctly with the candidate's value of $v$ .
The numerical value of $m$ is $0.3$ ; the unit kg may be included or omitted.
Keep directions consistent: all motion in this question is along the same straight line.

Techniques used

substitute result from part (a)apply conservation of linear momentum in a second collision

The rest of this paper

6 more questions

Q2Kinematics of Motion in a Straight Line · Energy, Work and Power6M
Q3Kinematics of Motion in a Straight Line4M
Q4Kinematics of Motion in a Straight Line8M
Q5Forces and Equilibrium9M
Q6Newton's Laws of Motion · Forces and Equilibrium8M
Q7Energy, Work and Power · Newton's Laws of Motion · Forces and Equilibrium11M

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