Mathematics 9709/42 — October/November 2020

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

questions

marks

minutes

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

Q1MomentumFree sample

Two particles $P$ and $Q$ , of masses $0.2\text{ kg}$ and $0.5\text{ kg}$ respectively, are at rest on a smooth horizontal plane. $P$ is projected towards $Q$ with speed $2\text{ m s}^{-1}$ .

(a)

Write down the momentum of $P$ .

DifficultyEasy

Worked solution

Approach

Use the definition of linear momentum: momentum is the product of mass and velocity.

Working

p = mv = 0.2 \times 2 = 0.4

The momentum of $P$ is $0.4\text{ kg m s}^{-1}$ in the direction of projection.

Answer

0.4\text{ kg m s}^{-1}

Final answer

0.4 kg m s^{-1}

Detailed explanation

Walkthrough

This part only asks for the momentum of $P$ before the collision. Momentum is defined as the product of mass and velocity, so multiply $0.2\text{ kg}$ by $2\text{ m s}^{-1}$ . Since the motion is in one horizontal direction, the momentum has magnitude $0.4\text{ kg m s}^{-1}$ and points in the direction of projection.

Key Takeaways

Momentum is a vector quantity; in one-dimensional motion we can treat it as a signed scalar. The unit is $\text{kg m s}^{-1}$ .

Common Mistakes

Forgetting the units of momentum.
Using weight instead of mass.
Omitting the direction when momentum is requested as a vector.

Things to Be Careful About

The plane being smooth is not needed for this part; momentum depends only on mass and velocity. The answer should be given with correct units.

Techniques used

apply the definition of linear momentummultiply mass by velocitystate momentum with correct units

(b)

After the collision $P$ continues to move in the same direction with speed $0.3\text{ m s}^{-1}$ .

Find the speed of $Q$ after the collision.

DifficultyMedium-Easy

Worked solution

Approach

Because the plane is smooth and horizontal, there is no external horizontal force during the collision, so total momentum is conserved. Use the initial momentum of $P$ from part (a) and set it equal to the sum of the final momenta of $P$ and $Q$ .

Working

Before the collision, only $P$ is moving, so the total momentum is:

0.4\text{ kg m s}^{-1}

After the collision, $P$ has momentum $0.2 \times 0.3$ and $Q$ has momentum $0.5v$ , where $v$ is the speed of $Q$ .

0.4 = 0.2 \times 0.3 + 0.5v

0.4 = 0.06 + 0.5v

0.5v = 0.34

v = 0.68\text{ m s}^{-1}

Answer

0.68\text{ m s}^{-1}

Final answer

0.68 m s^{-1}

Detailed explanation

Walkthrough

Start with the total momentum before the collision. $Q$ is at rest, so its momentum is zero and the total is just the momentum of $P$ , $0.4\text{ kg m s}^{-1}$ . After the collision, $P$ still moves in the same direction with speed $0.3\text{ m s}^{-1}$ , so its momentum is $0.2 \times 0.3 = 0.06$ . Let $v$ be the speed of $Q$ ; its momentum is $0.5v$ . Conservation of momentum gives $0.4 = 0.06 + 0.5v$ . Subtract $0.06$ from both sides to get $0.5v = 0.34$ , then divide by $0.5$ to get $v = 0.68\text{ m s}^{-1}$ .

Key Takeaways

In a collision on a smooth horizontal plane, momentum is conserved because no external horizontal force acts. The total momentum before equals the sum of momenta after. Choose a positive direction and use consistent signs.

Common Mistakes

Forgetting that $Q$ is initially at rest, so its initial momentum is zero.
Omitting the $0.5v$ term for $Q$ after the collision.
Using the final speed of $P$ as the total final momentum.
Not following through from part (a) if the initial momentum was calculated incorrectly.

Things to Be Careful About

Units: momentum in $\text{kg m s}^{-1}$ , speed in $\text{m s}^{-1}$ .
Direction: both particles move in the same direction after the collision, so all terms are positive with the chosen sign convention.
The mark scheme allows follow-through on the value from part (a).

Techniques used

apply conservation of linear momentumset initial momentum equal to sum of final momentasolve a linear equation for the unknown speed

The rest of this paper

7 more questions

Q2Newton's Laws of Motion · Energy, Work and Power5M
Q3Forces and Equilibrium5M
Q4Kinematics of Motion in a Straight Line5M
Q5Kinematics of Motion in a Straight Line8M
Q6Forces and Equilibrium8M
Q7Kinematics of Motion in a Straight Line7M
Q8Newton's Laws of Motion · Energy, Work and Power9M

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