Physics 9702 · Paper 1

Explosions & Recoil

Conservation of momentum also works in reverse: not two objects becoming one, but one object becoming two.
A nucleus decays, a gun fires, two skaters push apart. The same single idea solves all of them.

Total momentum zero, before and after

If the object starts at rest, the total momentum before is zero. So the total after must also be zero.
Two pieces flying apart must therefore carry equal-size momentum in opposite directions: $m_1 v_1 = m_2 v_2$ (sizes), directions opposite.
The light piece moves fast, the heavy piece moves slowly, in exactly the ratio that makes the momenta match.

Total momentum starts at zero, so it must stay zero: the two parts carry equal-size momentum in opposite directions. The light part moves fast, the heavy part moves slowly.

Two ice skaters gliding away from each other on smooth ice after pushing apart. — The two skaters start at rest, so the total momentum is zero. When they push apart, the total must stay zero, so they gain equal and opposite momenta. The lighter skater moves off faster, so that mass × velocity is the same size on each side.

Worked example

Smallest case: two skaters

Two skaters stand still on smooth ice, then push apart. The 60 kg skater moves off at 2.0 m s⁻¹. Find the velocity of the 40 kg skater.

Total momentum starts at zero, so: $40\,v = 60 \times 2.0$ .
$v = \dfrac{120}{40} = 3.0 \text{ m s}^{-1}$ in the opposite direction.

Answer

v = 3.0 \text{ m s}^{-1} \text{ opposite to the 60 kg skater}

Worked example

Exam version: alpha decay

A stationary nucleus decays into an alpha particle (mass 4u) and a nucleus Q (mass 215u). Q moves off at 3.2 × 10⁵ m s⁻¹. Find the speed of the alpha particle. (9702/22/F/M/24 Q4(b))

Total momentum is zero, so the sizes match: $4u \times v_\alpha = 215u \times 3.2 \times 10^{5}$ .
$v_\alpha = \dfrac{215 \times 3.2 \times 10^{5}}{4} = 1.7 \times 10^{7} \text{ m s}^{-1}$ .

Answer

v_\alpha = 1.7 \times 10^{7} \text{ m s}^{-1}

The explain version: “explain why the two parts move in exactly opposite directions.” Answer: momentum is conserved and the total starts at zero; if the parts moved at an angle, there would be a sideways momentum after the decay that was not there before.

Your turn— tap to reveal the worked answer (9702-style)

A rifle of mass 4.0 kg fires a bullet of mass 0.010 kg at 400 m s⁻¹. Find the recoil speed of the rifle.

$4.0\,v = 0.010 \times 400 = 4.0 \text{ kg m s}^{-1}$ .
$v = 1.0 \text{ m s}^{-1}$ backwards.

Answer: 1.0 m s⁻¹ backwards.

Starts at rest and splits apart: write $m_1 v_1 = m_2 v_2$ , directions opposite. That is the whole method.

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Explosions & Recoil

Conservation of momentum also works in reverse: not two objects becoming one, but one object becoming two.
A nucleus decays, a gun fires, two skaters push apart. The same single idea solves all of them.

Total momentum zero, before and after

If the object starts at rest, the total momentum before is zero. So the total after must also be zero.
Two pieces flying apart must therefore carry equal-size momentum in opposite directions: $m_1 v_1 = m_2 v_2$ (sizes), directions opposite.
The light piece moves fast, the heavy piece moves slowly, in exactly the ratio that makes the momenta match.

Total momentum starts at zero, so it must stay zero: the two parts carry equal-size momentum in opposite directions. The light part moves fast, the heavy part moves slowly.

Worked example

Smallest case: two skaters

Two skaters stand still on smooth ice, then push apart. The 60 kg skater moves off at 2.0 m s⁻¹. Find the velocity of the 40 kg skater.

Total momentum starts at zero, so: $40\,v = 60 \times 2.0$ .
$v = \dfrac{120}{40} = 3.0 \text{ m s}^{-1}$ in the opposite direction.

Answer

v = 3.0 \text{ m s}^{-1} \text{ opposite to the 60 kg skater}

Worked example

Exam version: alpha decay

A stationary nucleus decays into an alpha particle (mass 4u) and a nucleus Q (mass 215u). Q moves off at 3.2 × 10⁵ m s⁻¹. Find the speed of the alpha particle. (9702/22/F/M/24 Q4(b))

Total momentum is zero, so the sizes match: $4u \times v_\alpha = 215u \times 3.2 \times 10^{5}$ .
$v_\alpha = \dfrac{215 \times 3.2 \times 10^{5}}{4} = 1.7 \times 10^{7} \text{ m s}^{-1}$ .

Answer

v_\alpha = 1.7 \times 10^{7} \text{ m s}^{-1}

The explain version: “explain why the two parts move in exactly opposite directions.” Answer: momentum is conserved and the total starts at zero; if the parts moved at an angle, there would be a sideways momentum after the decay that was not there before.

Your turn— tap to reveal the worked answer (9702-style)

A rifle of mass 4.0 kg fires a bullet of mass 0.010 kg at 400 m s⁻¹. Find the recoil speed of the rifle.

$4.0\,v = 0.010 \times 400 = 4.0 \text{ kg m s}^{-1}$ .
$v = 1.0 \text{ m s}^{-1}$ backwards.

Answer: 1.0 m s⁻¹ backwards.

Starts at rest and splits apart: write $m_1 v_1 = m_2 v_2$ , directions opposite. That is the whole method.