Mathematics 9709/42 — February/March 2021

Walkthrough

We need to find how fast $P$ moves after colliding with $Q$. The key idea is that on a smooth horizontal plane there is no friction, so during the collision the total momentum of the two particles in the horizontal direction is conserved.

First choose a positive direction. It is convenient to take the direction in which $P$ is initially moving as positive. $P$'s initial velocity is then $+0.5$ m s$^{-1}$. $Q$ is moving towards $P$, so $Q$'s velocity is in the opposite direction; we write it as $-1$ m s$^{-1}$. This sign is essential: momentum is a vector quantity, so opposite directions must have opposite signs.

Before the collision, $P$'s momentum is $0.2 \times 0.5 = 0.1$ kg m s$^{-1}$. $Q$'s momentum is $0.3 \times (-1) = -0.3$ kg m s$^{-1}$. The total momentum before is $0.1 + (-0.3) = -0.2$ kg m s$^{-1}$.

After the collision, $Q$ comes to rest, so its momentum is $0$. Let $v$ be $P$'s velocity after the collision. Then $P$'s momentum after is $0.2v$. Conservation of momentum says:

Solving gives $-0.2 = 0.2v$, so $v = -1$ m s$^{-1}$. The negative sign means $P$ is now moving in the opposite direction to its original motion. Speed is the magnitude of velocity, so the speed is $1$ m s$^{-1}$.

Key Takeaways

• Momentum is conserved in a collision when no external horizontal force acts.
• Momentum is a vector: choose a positive direction and give velocities signs.
• Speed is the magnitude of velocity, so a negative velocity still gives a positive speed.
• Always include units (kg m s$^{-1}$ for momentum, m s$^{-1}$ for speed).

Common Mistakes

• Forgetting that $Q$ moves in the opposite direction and using $+1$ instead of $-1$ for $Q$'s velocity.
• Giving $v = -1$ m s$^{-1}$ as the final speed without taking the magnitude. The speed is $1$ m s$^{-1}$.
• Mixing units, e.g. using grams for mass in one term and kilograms in another.
• Omitting the zero momentum term for $Q$ after the collision, leading to an incorrect equation.

Things to Be Careful About

• The mark scheme allows either kg or g consistently, but the momentum equation must be dimensionally correct.
• The sign convention must be applied to every velocity term; the mark scheme allows three relevant momentum terms regardless of sign, but the final speed must come from a correct signed equation.
• If you choose the opposite positive direction, the signs of all velocities reverse, but the final speed remains $1$ m s$^{-1}$.
• The collision is direct and on a smooth plane, so only the two particles' momenta are involved; there is no impulse from the plane in the horizontal direction.