Newton’s laws

# Newton’s laws ## Objectifs et prérequis This chapter introduces the concept of linear momentum and its conservation in interactions, grounded in Newton’s laws of motion. Students will learn to define momentum, analyze collisions and explosions using conservation principles, and solve problems involving isolated systems. Prerequisites include understanding of Newton’s second law (F = ma), vector quantities, kinematics in one and two dimensions, and basic algebra and trigonometry. Familiarity with force diagrams and impulse is beneficial but will be reinforced. ## Introduction Linear momentum, a vector quantity defined as the product of an object’s mass and its velocity, provides a powerful framework for analyzing motion during interactions where forces may be complex or time-varying. Newton’s second law can be expressed in terms of momentum: the net force acting on a system equals the rate of change of its total momentum. When the net external force on a system is zero, the total momentum of the system remains constant—a principle known as conservation of momentum. This law is particularly useful in analyzing collisions (elastic, inelastic) and explosions, where internal forces dominate and external forces may be neglected over short time intervals. The conservation of momentum arises directly from Newton’s third law and is a cornerstone of classical mechanics, applicable from subatomic particles to celestial bodies. ## I. Linear momentum ### Definition and properties Linear momentum ($\vec{p}$) of a single particle is defined as the product of its mass ($m$) and its velocity ($\vec{v}$): > $$\vec{p} = m\vec{v}$$ Momentum is a vector quantity, possessing both magnitude and direction, aligned with the velocity vector. Its SI unit is kilogram-meter per second ($\text{kg} \cdot \text{m/s}$), equivalent to newton-second ($\text{N} \cdot \text{s}$). For a system of particles, the total linear momentum ($\vec{P}$) is the vector sum of the individual momenta: > $$\vec{P} = \sum_i \vec{p}_i = \sum_i m_i \vec{v}_i$$ If the mass distribution is continuous, the sum becomes an integral over the mass elements. Momentum is conserved in an isolated system (one with zero net external force), regardless of the nature of internal forces. ### Relationship with Newton’s second law Newton’s second law states that the net force ($\vec{F}_{\text{net}}$) acting on a particle equals the rate of change of its momentum: > $$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$$ For constant mass, this reduces to $\vec{F}_{\text{net}} = m\vec{a}$, since $\frac{d}{dt}(m\vec{v}) = m\frac{d\vec{v}}{dt} = m\vec{a}$. However, the momentum form is more general and remains valid even when mass changes (e.g., rockets), though such cases are beyond the scope of Form Five. The impulse-momentum theorem follows directly: the impulse ($\vec{J}$) delivered by a net force over a time interval $\Delta t$ equals the change in momentum: > $$\vec{J} = \int_{t_i}^{t_f} \vec{F}_{\text{net}} \, dt = \Delta \vec{p} = \vec{p}_f - \vec{p}_i$$ For constant average force, $\vec{J} = \vec{F}_{\text{avg}} \Delta t$. ### Examples **Example 1:** A 0.15 kg baseball is thrown horizontally at 40 m/s. What is its momentum? Solution: $\vec{p} = m\vec{v} = (0.15 \, \text{kg})(40 \, \text{m/s}) = 6.0 \, \text{kg} \cdot \text{m/s}$ in the direction of throw. **Example 2:** A 1200 kg car moving at 20 m/s east collides with a stationary 1800 kg truck. After collision, they stick together. What is their common velocity? Solution: This is an inelastic collision. Use conservation of momentum (covered in next section), but we compute initial momentum here: Initial momentum: $\vec{p}_i = (1200)(20) + (1800)(0) = 24,000 \, \text{kg} \cdot \text{m/s}$ east. Final momentum: $\vec{p}_f = (1200 + 1800)v_f = 3000 v_f$. Set equal: $3000 v_f = 24,000 \Rightarrow v_f = 8.0 \, \text{m/s}$ east. (Note: This anticipates Section II; momentum conservation justifies the approach.) ### Exercices d’application corrigés **Exercise 1:** A 0.020 kg bullet is fired from a rifle at 500 m/s. Calculate its momentum. Solution: $p = mv = (0.020)(500) = 10.0 \, \text{kg} \cdot \text{m/s}$. **Exercise 2:** A 70 kg person runs at 3.0 m/s north. What is the magnitude and direction of their momentum? Solution: $p = (70)(3.0) = 210 \, \text{kg} \cdot \text{m/s}$, direction north. **Exercise 3:** A 2.0 kg object has a momentum of 12.0 kg·m/s west. What is its velocity? Solution: $v = p/m = 12.0 / 2.0 = 6.0 \, \text{m/s}$ west. **Exercise 4:** A force of 50 N acts on a 5.0 kg object for 3.0 s. If the object starts from rest, what is its final momentum? Solution: Impulse $J = F \Delta t = (50)(3.0) = 150 \, \text{N} \cdot \text{s} = 150 \, \text{kg} \cdot \text{m/s}$. Initial momentum = 0, so final momentum = 150 kg·m/s in direction of force. ## II. Interaction and conservation of momentum ### Principle of conservation of momentum For an isolated system (one experiencing zero net external force), the total linear momentum remains constant over time: > $$\vec{P}_{\text{total, initial}} = \vec{P}_{\text{total, final}} \quad \text{or} \quad \sum_i m_i \vec{v}_{i,\text{initial}} = \sum_i m_i \vec{v}_{i,\text{final}}$$ This principle is a direct consequence of Newton’s third law: for every action force, there is an equal and opposite reaction force. Internal forces occur in equal-and-opposite pairs, so their contributions to the total momentum change cancel out. Only external forces can change the total momentum of a system. If $\vec{F}_{\text{ext, net}} = 0$, then $d\vec{P}/dt = 0$, so $\vec{P}$ is constant. ### Types of interactions Interactions are classified based on kinetic energy conservation: - **Elastic collision:** Both momentum and kinetic energy are conserved. Objects bounce apart without deformation or heat generation (idealized). - **Inelastic collision:** Momentum is conserved, but kinetic energy is not (some converts to internal energy: heat, sound, deformation). - **Perfectly inelastic collision:** Objects stick together after collision; maximum kinetic energy is lost while momentum is conserved. - **Explosion:** A single object separates into fragments; momentum is conserved (initial momentum zero if object was at rest), but kinetic energy increases (from internal potential energy). ### Solving problems: step-by-step method 1. **Define the system:** Identify all objects involved and confirm it is isolated (net external force ≈ 0 during interaction). 2. **Draw diagrams:** Sketch before and after interaction, labeling masses and velocities (use coordinate axes). 3. **Write momentum conservation equation:** Set total initial momentum equal to total final momentum. 4. **Solve for unknowns:** Use vector components if motion is in two dimensions. 5. **Check:** Verify units, direction, and plausibility (e.g., speeds not exceeding reasonable limits). ### Examples **Example 1 (Elastic collision):** A 0.500 kg ball moving at 4.00 m/s right strikes a stationary 1.00 kg ball elastically. Find velocities after collision. Solution: Let $m_1 = 0.500 \, \text{kg}$, $v_{1i} = 4.00 \, \text{m/s}$, $m_2 = 1.00 \, \text{kg}$, $v_{2i} = 0$. Conservation of momentum: $$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$$ $$(0.500)(4.00) + (1.00)(0) = 0.500 v_{1f} + 1.00 v_{2f}$$ $$2.00 = 0.500 v_{1f} + v_{2f} \quad \text{(1)}$$ Conservation of kinetic energy (elastic): $$\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2$$ $$\frac{1}{2}(0.500)(16.0) + 0 = \frac{1}{2}(0.500) v_{1f}^2 + \frac{1}{2}(1.00) v_{2f}^2$$ $$4.00 = 0.250 v_{1f}^2 + 0.500 v_{2f}^2 \quad \text{(2)}$$ From (1): $v_{2f} = 2.00 - 0.500 v_{1f}$. Substitute into (2): $$4.00 = 0.250 v_{1f}^2 + 0.500 (2.00 - 0.500 v_{1f})^2$$ $$4.00 = 0.250 v_{1f}^2 + 0.500 (4.00 - 2.00 v_{1f} + 0.250 v_{1f}^2)$$ $$4.00 = 0.250 v_{1f}^2 + 2.00 - 1.00 v_{1f} + 0.125 v_{1f}^2$$ $$4.00 = 0.375 v_{1f}^2 - 1.00 v_{1f} + 2.00$$ $$0 = 0.375 v_{1f}^2 - 1.00 v_{1f} - 2.00$$ Multiply by 8: $0 = 3 v_{1f}^2 - 8 v_{1f} - 16$ Solve quadratic: $v_{1f} = \frac{8 \pm \sqrt{64 + 192}}{6} = \frac{8 \pm \sqrt{256}}{6} = \frac{8 \pm 16}{6}$ Take negative root (since ball rebounds): $v_{1f} = \frac{8 - 16}{6} = -\frac{8}{6} = -1.33 \, \text{m/s}$ Then $v_{2f} = 2.00 - 0.500(-1.33) = 2.00 + 0.665 = 2.665 \approx 2.67 \, \text{m/s}$ Answer: $v_{1f} = -1.33 \, \text{m/s}$ (left), $v_{2f} = 2.67 \, \text{m/s}$ (right). **Example 2 (Perfectly inelastic collision):** A 1500 kg car moving at 12 m/s east hits a 2500 kg van moving at 8 m/s west. They stick together. Find final velocity. Solution: Let east be positive. $m_c = 1500 \, \text{kg}$, $v_{ci} = +12 \, \text{m/s}$ $m_v = 2500 \, \text{kg}$, $v_{vi} = -8 \, \text{m/s}$ (west) Initial momentum: $$p_i = m_c v_{ci} + m_v v_{vi} = (1500)(12) + (2500)(-8) = 18,000 - 20,000 = -2,000 \, \text{kg} \cdot \text{m/s}$$ Final mass: $m_f = 1500 + 2500 = 4000 \, \text{kg}$ Final momentum: $p_f = m_f v_f = 4000 v_f$ Conservation: $4000 v_f = -2000 \Rightarrow v_f = -0.50 \, \text{m/s}$ Answer: $0.50 \, \text{m/s}$ west. **Example 3 (Explosion):** A 5.0 kg shell at rest explodes into two fragments: 2.0 kg and 3.0 kg. The 2.0 kg fragment moves at 30 m/s north. Find velocity of 3.0 kg fragment. Solution: Initial momentum = 0 (at rest). Let north be positive. $m_1 = 2.0 \, \text{kg}$, $v_{1f} = +30 \, \text{m/s}$ $m_2 = 3.0 \, \text{kg}$, $v_{2f} = ?$ Conservation: $$0 = m_1 v_{1f} + m_2 v_{2f}$$ $$0 = (2.0)(30) + (3.0) v_{2f}$$ $$0 = 60 + 3.0 v_{2f}$$ $$v_{2f} = -20 \, \text{m/s}$$ Answer: $20 \, \text{m/s}$ south. ### Exercices d’application corrigés **Exercise 1:** A 0.10 kg hockey puck moving at 8.0 m/s east collides elastically with a stationary 0.20 kg puck. Find velocities after collision. Solution: $m_1 = 0.10$, $v_{1i} = 8.0$; $m_2 = 0.20$, $v_{2i} = 0$ Momentum: $0.10 \times 8.0 = 0.10 v_{1f} + 0.20 v_{2f} \Rightarrow 0.80 = 0.10 v_{1f} + 0.20 v_{2f}$ → (1) KE: $\frac{1}{2}(0.10)(64) = \frac{1}{2}(0.10) v_{1f}^2 + \frac{1}{2}(0.20) v_{2f}^2 \Rightarrow 3.2 = 0.05 v_{1f}^2 + 0.10 v_{2f}^2$ → (2) From (1): $v_{2f} = (0.80 - 0.10 v_{1f})/0.20 = 4.0 - 0.5 v_{1f}$ Plug into (2): $3.2 = 0.05 v_{1f}^2 + 0.10 (4.0 - 0.5 v_{1f})^2$ $3.2 = 0.05 v_{1f}^2 + 0.10 (16.0 - 4.0 v_{1f} + 0.25 v_{1f}^2)$ $3.2 = 0.05 v_{1f}^2 + 1.6 - 0.4 v_{1f} + 0.025 v_{1f}^2$ $3.2 = 0.075 v_{1f}^2 - 0.4 v_{1f} + 1.6$ $1.6 = 0.075 v_{1f}^2 - 0.4 v_{1f}$ Multiply by 40: $64 = 3 v_{1f}^2 - 16 v_{1f}$ $3 v_{1f}^2 - 16 v_{1f} - 64 = 0$ $v_{1f} = \frac{16 \pm \sqrt{256 + 768}}{6} = \frac{16 \pm \sqrt{1024}}{6} = \frac{16 \pm 32}{6}$ Take negative: $v_{1f} = (16 - 32)/6 = -16/6 = -2.67 \, \text{m/s}$ $v_{2f} = 4.0 - 0.5(-2.67) = 4.0 + 1.335 = 5.335 \approx 5.33 \, \text{m/s}$ Answer: $v_{1f} = -2.67 \, \text{m/s}$ (west), $v_{2f} = 5.33 \, \text{m/s}$ (east). **Exercise 2:** A 900 kg car rear-ends a 1200 kg stationary truck. After collision, they move together at 5.0 m/s east. What was the car’s speed before collision? Solution: Let east be positive. $m_c = 900$, $v_{ci} = ?$; $m_t = 1200$, $v_{ti} = 0$; $v_f = 5.0$ Initial momentum: $900 v_{ci} + 1200 \times 0 = 900 v_{ci}$ Final momentum: $(900 + 1200) \times 5.0 = 2100 \times 5.0 = 10,500$ Conservation: $900 v_{ci} = 10,500 \Rightarrow v_{ci} = 11.67 \, \text{m/s}$ east. **Exercise 3:** A 4.0 kg object moving at 5.0 m/s right explodes into two 2.0 kg fragments. One moves at 3.0 m/s left. Find velocity of the other. Solution: Initial momentum: $p_i = (4.0)(5.0) = 20.0 \, \text{kg} \cdot \text{m/s}$ right (+) Let right be positive. Fragment 1: $m_1 = 2.0$, $v_{1f} = -3.0$ (left) Fragment 2: $m_2 = 2.0$, $v_{2f} = ?$ Final momentum: $p_f = (2.0)(-3.0) + (2.0) v_{2f} = -6.0 + 2.0 v_{2f}$ Conservation: $-6.0 + 2.0 v_{2f} = 20.0$ $2.0 v_{2f} = 26.0 \Rightarrow v_{2f} = 13.0 \, \text{m/s}$ right. **Exercise 4:** Two ice skaters, masses 50 kg and 70 kg, push off from rest. If the 50 kg skater moves at 3.0 m/s, what is the velocity of the 70 kg skater? Solution: Initial momentum = 0. Let direction of 50 kg skater be positive. $0 = (50)(3.0) + (70) v_{2f}$ $0 = 150 + 70 v_{2f}$ $v_{2f} = -150/70 = -2.14 \, \text{m/s}$ Answer: $2.14 \, \text{m/s}$ opposite to the 50 kg skater. ## Synthèse du chapitre - Linear momentum ($\vec{p} = m\vec{v}$) is a vector quantity conserved in isolated systems. - Newton’s second law in momentum form: $\vec{F}_{\text{net}} = d\vec{p}/dt$; impulse-momentum theorem: $\vec{J} = \Delta \vec{p}$. - For a system: $\vec{P}_{\text{total}} = \sum m_i \vec{v}_i$; conserved if $\vec{F}_{\text{ext, net}} = 0$. - Collisions: elastic (KE conserved), inelastic (KE not conserved), perfectly inelastic (objects stick). - Explosions: initial momentum zero if object at rest; fragments fly apart with net momentum zero. - Problem-solving: isolate system, draw diagrams, apply $\sum \vec{p}_i = \sum \vec{p}_f$, solve components. - Conservation of momentum follows from Newton’s third law; internal forces cancel in pairs. - Applicable to 1D and 2D motion; vector addition essential for oblique collisions. - Units: momentum in $\text{kg} \cdot \text{m/s}$ or $\text{N} \cdot \text{s}$; impulse identical dimensionally. ## Glossaire - **Linear momentum**: Product of an object’s mass and its velocity ($\vec{p} = m\vec{v}$); vector quantity. - **Impulse**: Change in momentum due to a net force over time ($\vec{J} = \int \vec{F} \, dt$); equals $\Delta \vec{p}$. - **Isolated system**: A system experiencing zero net external force ($\vec{F}_{\text{ext, net}} = 0$). - **Conservation of momentum**: Principle that total momentum of an isolated system remains constant. - **Elastic collision**: Interaction where both momentum and kinetic energy are conserved. - **Inelastic collision**: Interaction where momentum is conserved but kinetic energy is not. - **Perfectly inelastic collision**: Collision in which objects stick together, maximizing kinetic energy loss. - **Explosion**: Process where a single object separates into fragments; momentum conserved if initial momentum zero. - **Internal forces**: Forces exerted between objects within a system; occur in action-reaction pairs. - **External forces**: Forces exerted on a system by objects outside it; can change total momentum. - **Kinetic energy**: Energy of motion ($\frac{1}{2}mv^2$); not necessarily conserved in collisions. - **Center of mass**: Point representing average position of mass in a system; moves as if all external forces act there.