Conservation of momentum in two dimensions
Two cars meet at a junction outside Mullingar: one is heading east, the other north. They crumple together and slide off in a direction that is neither east nor north. How can we predict where the wreckage goes, and how fast? In a straight line, conservation of momentum was one equation. In a plane, momentum is a vector, so we need two.
Momentum as a vector
Momentum is $p = mv$. Since velocity is a vector, so is momentum, and it points in the same direction as the velocity. Its SI unit is $\mathrm{kg\,m\,s^{-1}}$.
To work in a plane, choose perpendicular axes $x$ and $y$ and resolve the velocity into components. If a body of mass $m$ moves at speed $v$ at angle $\theta$ above the $x$-axis:
$$p_x = mv\cos\theta \qquad p_y = mv\sin\theta$$To get back to the magnitude and direction, use Pythagoras and trigonometry:
$$p = \sqrt{p_x^2 + p_y^2} \qquad \tan\theta = \frac{p_y}{p_x}$$A ball of mass $0.50\ \text{kg}$ moves with velocity components $v_x = 4.0\ \text{m s}^{-1}$ and $v_y = 3.0\ \text{m s}^{-1}$. a) Calculate the speed of the ball. b)…
A ball of mass $0.50\ \text{kg}$ moves with velocity components $v_x = 4.0\ \text{m s}^{-1}$ and $v_y = 3.0\ \text{m s}^{-1}$.
- a) Calculate the speed of the ball.
- b) Calculate the $x$ and $y$ components of its momentum.
- c) Calculate the magnitude of its momentum.
A $1,000\ \text{kg}$ car travelling east at $10\ \text{m s}^{-1}$ and a $1,500\ \text{kg}$ van travelling north at $8.0\ \text{m s}^{-1}$ collide at a junction in…
A $1,000\ \text{kg}$ car travelling east at $10\ \text{m s}^{-1}$ and a $1,500\ \text{kg}$ van travelling north at $8.0\ \text{m s}^{-1}$ collide at a junction in Mullingar.
- a) Calculate the momentum of the car and of the van, giving the direction of each.
- b) Calculate the $x$ and $y$ components of the total momentum before the collision, taking east as $+x$ and north as $+y$.
- c) State the principle you used to describe the total momentum after the collision, and the condition under which it applies.
The principle in two dimensions
Principle of conservation of momentum: in any collision or interaction between bodies, the total momentum of the system before equals the total momentum after, provided no external resultant force acts on the system.
Newton's third law is the reason: during the collision the two bodies exert equal and opposite forces on each other for the same time, so the changes in momentum are equal and opposite. The total cannot change.
Because momentum is a vector, the principle holds in each direction separately:
$$\sum p_x\ (\text{before}) = \sum p_x\ (\text{after}) \qquad \sum p_y\ (\text{before}) = \sum p_y\ (\text{after})$$Two equations means we can find two unknowns, for example the speed and direction of the second body.
Two snooker balls on a table in a club in Navan have equal masses of $0.16\ \text{kg}$. A ball moving at $2.0\ \text{m s}^{-1}$ along the $x$-axis strikes a stationary…
Two snooker balls on a table in a club in Navan have equal masses of $0.16\ \text{kg}$. A ball moving at $2.0\ \text{m s}^{-1}$ along the $x$-axis strikes a stationary ball. After the collision the first ball moves at $1.2\ \text{m s}^{-1}$ at $+53^\circ$ to the $x$-axis, and the second moves off at an angle below the $x$-axis.
- a) Calculate the total momentum in the $x$-direction before the collision.
- b) State the total momentum in the $y$-direction before the collision.
- c) State what the total $y$-momentum must be after the collision, and explain why the second ball must move below the axis.
Perfectly inelastic collisions
In an inelastic collision, momentum is conserved but kinetic energy is not: some is converted to heat, sound and deformation. If the bodies stick together, the collision is perfectly inelastic and they move with one common velocity.
A $1,000\ \text{kg}$ car travelling east at $10\ \text{m s}^{-1}$ and a $1,500\ \text{kg}$ van travelling north at $8.0\ \text{m s}^{-1}$ collide at a junction in…
A $1,000\ \text{kg}$ car travelling east at $10\ \text{m s}^{-1}$ and a $1,500\ \text{kg}$ van travelling north at $8.0\ \text{m s}^{-1}$ collide at a junction in Mullingar and stick together.
- a) Calculate the components of the velocity of the wreckage immediately after the collision.
- b) Calculate the speed of the wreckage and its direction, measured north of east.
- c) Calculate the kinetic energy lost in the collision.
At a cross-roads outside Ennistymon, a car of mass $1,200\ \text{kg}$ travelling east collides with a van of mass $2,000\ \text{kg}$ travelling north. They lock together…
At a cross-roads outside Ennistymon, a car of mass $1,200\ \text{kg}$ travelling east collides with a van of mass $2,000\ \text{kg}$ travelling north. They lock together and skid along a straight line at $35^\circ$ east of north, a distance of $14\ \text{m}$ before stopping. The coefficient of kinetic friction between the tyres and the road is $0.60$. Take $g = 9.8\ \text{m s}^{-2}$.
- a) Calculate the speed of the wreckage immediately after the collision.
- b) Calculate the momentum components of the wreckage after the collision.
- c) Calculate the initial speeds of the car and of the van.
- d) The speed limit on the road is $80\ \text{km h}^{-1}$. State, with a calculation, whether the car was speeding.
Elastic collisions and the right angle
In an elastic collision both momentum and kinetic energy are conserved. Check by calculating $E_k$ before and after: if they match, the collision is elastic.
In the lab we investigate conservation of momentum using two trolleys on a runway (or an air track) with light gates. A light gate times a card of known length passing through it, so $v = \text{card length}/\text{time}$ before and after the collision. Compare the total momentum before and after: within experimental error they agree, which verifies the principle. This is Newton's third law in action: the two trolleys exert equal and opposite forces for the same time, so their changes in momentum are equal and opposite.
Equal masses
When a moving body strikes a stationary body of equal mass in an elastic collision, the two move off at $90^\circ$ to each other.
Proof: let the mass be $m$, the initial speed $u$, and the final velocities $\vec v_1$ and $\vec v_2$.
Momentum: $m\vec u = m\vec v_1 + m\vec v_2$, so $\vec u = \vec v_1 + \vec v_2$.
Squaring (dot product with itself): $u^2 = v_1^2 + v_2^2 + 2\vec v_1\cdot\vec v_2$.
Energy: $\tfrac12 mu^2 = \tfrac12 mv_1^2 + \tfrac12 mv_2^2$, so $u^2 = v_1^2 + v_2^2$.
Comparing, $\vec v_1\cdot\vec v_2 = 0$, so the velocities are perpendicular: $\theta_1 + \theta_2 = 90^\circ$.
The vector triangle of momenta then has a right angle, and Pythagoras applies: the sum of the squares of the two shorter sides equals the square of the longest side.
A $0.40\ \text{kg}$ puck on an air table moves at $3.0\ \text{m s}^{-1}$ in the $+x$ direction and strikes a stationary $0.60\ \text{kg}$ puck. After the collision the…
A $0.40\ \text{kg}$ puck on an air table moves at $3.0\ \text{m s}^{-1}$ in the $+x$ direction and strikes a stationary $0.60\ \text{kg}$ puck. After the collision the $0.40\ \text{kg}$ puck moves at $1.0\ \text{m s}^{-1}$ at $60^\circ$ above the $+x$ axis.
- a) Calculate the $x$ and $y$ components of the velocity of the $0.60\ \text{kg}$ puck after the collision.
- b) Calculate the speed and direction of the $0.60\ \text{kg}$ puck.
- c) Determine whether the collision is elastic by calculating the kinetic energy before and after.
In an investigation of an elastic collision on a frictionless surface, a puck of mass $m$ moving at speed $u$ strikes a stationary puck of identical mass $m$. After the…
In an investigation of an elastic collision on a frictionless surface, a puck of mass $m$ moving at speed $u$ strikes a stationary puck of identical mass $m$. After the collision the pucks move off at angles $\theta_1$ and $\theta_2$ on opposite sides of the original line of motion.
- a) Using conservation of momentum, write equations for the $x$ and $y$ directions.
- b) By using conservation of kinetic energy and the vector triangle of momenta, show that $\theta_1+\theta_2=90^\circ$.
- c) In one trial $u = 2.0\ \text{m s}^{-1}$ and $\theta_1 = 30^\circ$. Calculate the speeds of both pucks after the collision.
- d) A student measures $\theta_1 = 30^\circ$ and $\theta_2 = 55^\circ$. State what this suggests about the collision and give two possible reasons.
A neutron of mass $m$ and speed $v$ collides elastically with a stationary nucleus of mass $4m$ and is scattered through $90^\circ$ from its original direction. a) Write…
A neutron of mass $m$ and speed $v$ collides elastically with a stationary nucleus of mass $4m$ and is scattered through $90^\circ$ from its original direction.
- a) Write equations for conservation of momentum in the directions parallel and perpendicular to the original motion of the neutron.
- b) Using conservation of kinetic energy, find the speed of the neutron after the collision, in terms of $v$.
- c) Find the speed of the nucleus after the collision and the angle at which it recoils to the original direction.
- d) Calculate the fraction of the neutron's initial kinetic energy transferred to the nucleus, and explain why light nuclei are better moderators.
Explosions and decays
If a stationary body breaks into pieces, the total momentum before is zero, so the vector sum of the momenta of the pieces must also be zero. This is why a gun recoils and why a radioactive nucleus at rest recoils when it emits particles.
With two pieces, the momenta are equal and opposite. With three, the momentum vectors form a closed triangle. In components: $\sum p_x = 0$ and $\sum p_y = 0$.
A stationary radioactive nucleus of mass $3.0\times10^{-25}\ \text{kg}$ decays and emits two particles. An alpha particle of mass $6.6\times10^{-27}\ \text{kg}$ moves…
A stationary radioactive nucleus of mass $3.0\times10^{-25}\ \text{kg}$ decays and emits two particles. An alpha particle of mass $6.6\times10^{-27}\ \text{kg}$ moves off at $1.5\times10^{6}\ \text{m s}^{-1}$ along the $+x$ axis. A second particle of mass $1.0\times10^{-26}\ \text{kg}$ moves off along the $+y$ axis at $2.0\times10^{6}\ \text{m s}^{-1}$. The remaining daughter nucleus recoils.
- a) State the total momentum before the decay.
- b) Calculate the components of momentum of the daughter nucleus.
- c) Calculate the magnitude of its momentum and the direction of its recoil relative to the $+x$ axis.
Summary
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Momentum $p = mv$ is a vector. Resolve it: $p_x = mv\cos\theta$, $p_y = mv\sin\theta$.
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In a collision or explosion with no external resultant force, total momentum is conserved separately in the $x$ and $y$ directions.
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Add momenta by components, then use $p = \sqrt{p_x^2 + p_y^2}$ and $\tan\theta = p_y/p_x$.
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Elastic: momentum and kinetic energy conserved. Inelastic: only momentum; bodies that stick together move with a common velocity.
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Equal masses, elastic, one initially at rest: the bodies separate at $90^\circ$.
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Back to the start: at the Mullingar junction the wreckage moves $50^\circ$ north of east at about $6.2\ \mathrm{m\,s^{-1}}$.