Mathematics 9709/41 — October/November 2020

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

questions

marks

minutes

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

Q1MomentumEnergy, Work and PowerFree sample

A particle $B$ of mass $5\text{ kg}$ is at rest on a smooth horizontal table. A particle $A$ of mass $2.5\text{ kg}$ moves on the table with a speed of $6\text{ m s}^{-1}$ and collides directly with $B$ . In the collision the two particles coalesce.

(a)

Find the speed of the combined particle after the collision.

DifficultyMedium-Easy

Worked solution

Approach

The table is smooth, so no external horizontal force acts during the collision and linear momentum is conserved. Because the particles coalesce, they move off together with a common speed $v$ after the impact.

Working

Before the collision only particle $A$ moves, so the total momentum is

2.5 \times 6 = 15\ \text{kg m s}^{-1}

Particle $B$ is at rest and contributes zero momentum. After the collision the combined mass is $2.5 + 5 = 7.5\ \text{kg}$ , moving with common speed $v$ :

\text{momentum after} = (2.5 + 5)v = 7.5v

Conserving momentum:

\begin{aligned} 7.5v &= 15 \\ v &= \frac{15}{7.5} = 2\ \text{m s}^{-1} \end{aligned}

Answer

The speed of the combined particle is

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

Final answer

2 m/s

Detailed explanation

Walkthrough

The table is described as smooth and horizontal, so there is no friction and during the brief interval of the collision no external horizontal force acts on the system of the two particles. Consequently the total horizontal momentum of the system is conserved. Before the collision only particle $A$ is moving, so its momentum is $2.5 \times 6 = 15\ \text{kg m s}^{-1}$ , while particle $B$ is at rest and contributes zero. Since the particles coalesce, they stick together and move as a single object of mass $2.5 + 5 = 7.5\ \text{kg}$ with one common speed $v$ . Equating total momentum before and after gives $7.5v = 15$ , and dividing by $7.5$ gives $v = 2\ \text{m s}^{-1}$ . This is the required speed of the combined particle.

Key Takeaways

Linear momentum is conserved during a collision whenever the system is isolated from external horizontal forces (here guaranteed by the smooth table).
A coalescing (perfectly inelastic) collision means both particles travel together with a single common final velocity.
The total momentum of a system is the algebraic sum of the momenta of its parts.

Common Mistakes

Forgetting that $B$ is initially at rest, so its initial momentum is zero.
Using the combined mass on the wrong side of the momentum equation.
Failing to recognise that coalescing requires a single common final speed $v$ for both particles.

Things to Be Careful About

Keep units consistent: mass in kg and speed in $\text{m s}^{-1}$ , so momentum is in $\text{kg m s}^{-1}$ .
This is a direct (head-on) collision, so both particles move along the same straight line and the vector nature of momentum reduces to a signed scalar equation.
Show the full conservation equation $2.5 \times 6 = 2.5v + 5v$ to earn the method mark.

Techniques used

apply conservation of linear momentum in a direct impactuse the coalesce condition to set a common final velocitysolve 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 of the system immediately before and immediately after the collision using $KE = \frac{1}{2}mv^2$ , then subtract the after-collision value from the before-collision value to find the loss. Before the impact only $A$ moves; afterwards the combined mass moves with the speed from part (a).

Working

Kinetic energy before the collision (only $A$ moving):

KE_{\text{before}} = \frac{1}{2} \times 2.5 \times 6^2 = \frac{1}{2} \times 2.5 \times 36 = 45\ \text{J}

Kinetic energy after the collision (combined mass $7.5\ \text{kg}$ at $v = 2\ \text{m s}^{-1}$ ):

KE_{\text{after}} = \frac{1}{2} \times 7.5 \times 2^2 = \frac{1}{2} \times 7.5 \times 4 = 15\ \text{J}

Loss of kinetic energy:

\begin{aligned} \text{loss} &= KE_{\text{before}} - KE_{\text{after}} \\ &= 45 - 15 = 30\ \text{J} \end{aligned}

Answer

The loss of kinetic energy due to the collision is

30\ \text{J}

Final answer

30 J

Detailed explanation

Walkthrough

Kinetic energy is a scalar given by $KE = \frac{1}{2}mv^2$ . Before the collision only particle $A$ is moving, so $KE_{\text{before}} = \frac{1}{2} \times 2.5 \times 6^2 = 45\ \text{J}$ . After the collision the combined mass $7.5\ \text{kg}$ moves with speed $v = 2\ \text{m s}^{-1}$ from part (a), so $KE_{\text{after}} = \frac{1}{2} \times 7.5 \times 2^2 = 15\ \text{J}$ . The loss of kinetic energy is the difference between these values: $45 - 15 = 30\ \text{J}$ . This energy is not destroyed but is converted into heat, sound and deformation energy during the inelastic impact.

Key Takeaways

$KE = \frac{1}{2}mv^2$ ; because kinetic energy is a scalar, the total KE of a system is the ordinary sum of the KE of each moving body.
In an inelastic (coalescing) collision kinetic energy is not conserved; some of it is always lost.
The loss of KE is found as the before-collision total minus the after-collision total.

Common Mistakes

Using the combined mass $7.5\ \text{kg}$ for the before-collision calculation; only $A$ is moving before the collision.
Using $mv^2$ instead of $\frac{1}{2}mv^2$ .
Subtracting the wrong way round and reporting a negative loss.
Failing to carry forward the correct speed $v$ from part (a).

Things to Be Careful About

The mark scheme credits the use of $KE = \frac{1}{2}mv^2$ and requires both the before and after kinetic energies to be correct; follow-through on $v$ from part (a) is allowed.
The final loss must be positive, since kinetic energy always decreases in an inelastic collision.
Give the answer in joules (J).

Techniques used

compute kinetic energy as half the mass times the square of the speedevaluate the total kinetic energy before and after the collisionsubtract the after-collision kinetic energy from the before-collision value to find the loss

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