Mathematics 9709/42 — May/June 2022

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

questions

marks

minutes

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

Q1MomentumEnergy, Work and PowerFree sample

Small smooth spheres $A$ and $B$ , of equal radii and of masses $5\text{ kg}$ and $3\text{ kg}$ respectively, lie on a smooth horizontal plane. Initially $B$ is at rest and $A$ is moving towards $B$ with speed $8.5\text{ m s}^{-1}$ . The spheres collide and after the collision $A$ continues to move in the same direction but with a quarter of the speed of $B$ .

(a)

Find the speed of $B$ after the collision.

DifficultyMedium-Easy

Worked solution

Approach

Use conservation of linear momentum along the direction of motion. Since the collision is direct and the plane is smooth, the total momentum before equals the total momentum after. Let $v$ be the speed of $B$ after the collision; then $A$ moves at $\frac{1}{4}v$ .

Working

Let the direction of $A$ 's initial motion be positive.

Before collision:

\text{momentum} = 5 \times 8.5

After collision:

\text{momentum} = 5 \times \frac{1}{4}v + 3v

Conservation of momentum gives:

\begin{aligned} 5 \times 8.5 &= 5 \times \frac{1}{4}v + 3v \\ 42.5 &= 1.25v + 3v \\ 42.5 &= 4.25v \\ v &= 10 \end{aligned}

Answer

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

Final answer

10 m s^-1

Detailed explanation

Walkthrough

We are told the spheres are smooth and on a smooth horizontal plane, so there is no friction or external horizontal force during the collision. That means the total horizontal momentum of the two spheres is conserved. Choose the direction of A's initial motion as positive. Let the speed of B after the collision be $v$ . The key piece of information is that A moves at a quarter of B's speed, so A's speed after the collision is $\frac{v}{4}$ , not $\frac{8.5}{4}$ . Write the momentum before as $5 \times 8.5$ . Write the momentum after as $5 \times \frac{v}{4} + 3v$ . Set them equal and solve: $42.5 = 1.25v + 3v = 4.25v$ , so $v = 10$ . Since the question asks for speed, the answer is positive $10\text{ m s}^{-1}$ .

Key Takeaways

Conservation of linear momentum applies when no external force acts in the direction of motion. In direct impact problems, set up total momentum before = total momentum after. Use the given relationship between the post-collision speeds to reduce the problem to one unknown.

Common Mistakes

Forgetting to include both spheres in the after-collision momentum.
Using A's post-collision speed as $\frac{8.5}{4}$ instead of $\frac{v}{4}$ . The statement says A has a quarter of B's speed, so if B's speed is $v$ , A's speed is $\frac{v}{4}$ .
Giving $-10$ for speed; speed is a scalar and must be non-negative. The mark scheme awards A0 for $-10$ .
Using equal final speeds; the mark scheme gives M0 if $|X|$ and $|Y|$ are subsequently used as equal.

Things to Be Careful About

Use consistent units: kg and m s $^{-1}$ give momentum in kg m s $^{-1}$ .
The method mark requires three momentum terms; show $5 \times 8.5 = 5 \times \frac{v}{4} + 3v$ clearly.
The answer must be a speed, so write 10, not $-10$ .

Techniques used

apply conservation of linear momentumuse the given relationship between post-collision speedssolve a linear equation for the unknown speed

(b)

Find the loss of kinetic energy of the system due to the collision.

DifficultyMedium-Easy

Worked solution

Approach

Calculate the total kinetic energy before and after the collision using $\text{KE} = \frac{1}{2}mv^2$ , then subtract the final total from the initial total. The difference is the kinetic energy lost in the collision.

Working

From part (a), after the collision:

v_B = 10\text{ m s}^{-1}, \quad v_A = \frac{1}{4} \times 10 = 2.5\text{ m s}^{-1}

Kinetic energy before:

\text{KE}_{\text{before}} = \frac{1}{2} \times 5 \times 8.5^2 = 180.625\text{ J}

Kinetic energy after:

\begin{aligned} \text{KE}_{\text{after}} &= \frac{1}{2} \times 5 \times 2.5^2 + \frac{1}{2} \times 3 \times 10^2 \\ &= 15.625 + 150 \\ &= 165.625\text{ J} \end{aligned}

Loss of kinetic energy:

\text{Loss} = 180.625 - 165.625 = 15\text{ J}

Answer

The loss of kinetic energy due to the collision is $15\text{ J}$ .

Final answer

15 J

Detailed explanation

Walkthrough

The kinetic energy of a moving object is $\frac{1}{2}mv^2$ . Before the collision only A is moving, so the total kinetic energy is $\frac{1}{2} \times 5 \times 8.5^2 = 180.625\text{ J}$ . After the collision, A moves at $2.5\text{ m s}^{-1}$ (one quarter of B's $10\text{ m s}^{-1}$ ) and B moves at $10\text{ m s}^{-1}$ . So the total kinetic energy after is $\frac{1}{2} \times 5 \times 2.5^2 + \frac{1}{2} \times 3 \times 10^2 = 15.625 + 150 = 165.625\text{ J}$ . The loss is the difference: $180.625 - 165.625 = 15\text{ J}$ . The collision is not perfectly elastic because kinetic energy is lost.

Key Takeaways

Kinetic energy is a scalar and is always non-negative. In a collision, total momentum is conserved but total kinetic energy is not necessarily conserved. To find energy loss, calculate total KE before and total KE after and subtract.

Common Mistakes

Using the combined mass $5 + 3 = 8\text{ kg}$ with one speed. The two spheres have different speeds after the collision, so you must add their separate kinetic energies.
Using $2.5\text{ m s}^{-1}$ for B or $10\text{ m s}^{-1}$ for A; from part (a), B has $10$ and A has $2.5$ .
Using $\frac{1}{2} \times (5+3) \times 8.5^2$ or $\frac{1}{2} \times (5+3) \times 2.5^2$ ; the mark scheme explicitly disallows these.
Forgetting to subtract: the question asks for loss, so the final answer is $180.625 - 165.625 = 15\text{ J}$ .

Things to Be Careful About

Use the speed from part (a); if part (a) was wrong, follow-through marks may be available for the energy calculation using their speeds.
Keep units: J = kg m $^2$ s $^{-2}$ .
The mark scheme accepts AWRT $\pm 15.0$ , so a small rounding difference is fine.

Techniques used

calculate initial kinetic energycalculate final kinetic energy of both spheressubtract to find the energy loss

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