Explosions
Newton's Laws & Momentum - OCR A-Level Physics
Key Definition
Explosion
A single stationary object separates into two or more pieces under the action of internal forces (chemical, nuclear, or stored elastic potential energyEnergy stored in the configuration of a system; can be gravitational, elastic, chemical, electrical or nuclear.). Total momentum is conserved; total kinetic energy INCREASES from zero because energy is released from internal stores.
A single stationary object separates into two or more pieces under the action of internal forces (chemical, nuclear, or stored elastic potential energyEnergy stored in the configuration of a system; can be gravitational, elastic, chemical, electrical or nuclear.). Total momentum is conserved; total kinetic energy INCREASES from zero because energy is released from internal stores.
$$0 = m_1 v_1 + m_2 v_2$$
- Total momentum before the explosion is zero (everything at rest). By conservation of momentumIn a closed system, total momentum before equals total momentum after., total momentum after must also be zero.
- The two pieces move in opposite directions with equal magnitudes of momentum: $m_1 v_1 = - m_2 v_2$.
- The lighter piece moves faster than the heavier piece (their speeds are in inverse ratio to their masses).
- Worked example: a 4.0 kg cannon fires a 0.020 kg ball at $200 \; \text{m s}^{-1}$. Conservation gives $0 = (4.0) v_{\text{cannon}} + (0.020)(200)$, so $v_{\text{cannon}} = -1.0 \; \text{m s}^{-1}$ (recoil in the opposite direction).
- Kinetic energy is NOT conserved. The KE after the explosion comes from chemical (cannon), nuclear (radioactive decay), or elastic (spring release) energy stored in the system before the event.
Worked Example [3 marks]
A 4.0 kg cannon fires a 0.020 kg ball at $200 \; \text{m s}^{-1}$. Calculate the recoil velocity of the cannon.
Show Solution
1
Total momentum before $= 0$ (both at rest). Conservation: $0 = m_{\text{c}} v_{\text{c}} + m_{\text{b}} v_{\text{b}}$.
[1]2
$0 = (4.0) v_{\text{c}} + (0.020)(200) = 4.0 \, v_{\text{c}} + 4.0$
[1]3
$v_{\text{c}} = -1.0 \; \text{m s}^{-1}$. The negative sign shows the cannon recoils in the direction opposite to the ball.
[1]Answer
$1.0 \; \text{m s}^{-1}$ in the opposite direction to the ball.
Common Mistake
MEDIUM
Wrong: Writing that "kinetic energy is conserved in an explosion" because energy must be conserved.
Right: TOTAL energy is conserved, but KINETIC energy is NOT. Before the explosion, KE is zero. After the explosion, both pieces have KE. The extra KE comes from chemical, elastic or nuclear potential energy that was stored in the object beforehand.
Right: TOTAL energy is conserved, but KINETIC energy is NOT. Before the explosion, KE is zero. After the explosion, both pieces have KE. The extra KE comes from chemical, elastic or nuclear potential energy that was stored in the object beforehand.
Examiner Tips and Tricks
- Always state where the kinetic energy came from: chemical store in the gunpowder, elastic PE in a compressed spring, nuclear potential in a decaying nucleus.
- A negative velocity is the expected answer for the heavier piece. Do not "fix" the sign; explain it as recoil.
- When asked to derive the recoil velocity formula, write $0 = m_1 v_1 + m_2 v_2$, then $v_2 = -m_1 v_1 / m_2$. The minus sign is the physically meaningful part of the answer.
Before/after sketch: cannon and ball at rest with internal spring/charge; after, ball moving right with long arrow and cannon moving left with short arrow. Equal-magnitude momentum vectors marked.