Linear momentum $\vec{p}$ is a vector quantity defined as the product of an object's mass $m$ and its velocity $\vec{v}$. It quantifies the inertia of a moving object. The standard SI unit is $\text{kg} \cdot \text{m/s}$.
The Impulse-Momentum Theorem states that the impulse $\vec{J}$ applied to an object is equal to the resulting change in the object's momentum $\Delta \vec{p}$.
Derivation from Newton's Second Law:
We define the net force $\vec{F}_{net}$ as the time rate of change of momentum:
Integrating both sides over the duration of the force, from an initial time $t_i$ to a final time $t_f$:
The left side is the Impulse $\vec{J}$ (unit $\text{N} \cdot \text{s}$), and the right side is the definite integral of the momentum change:
If the force is constant, the impulse simplifies to $\vec{J} = \vec{F}_{net} \Delta t$.
The relationship between kinetic energy ($K$) and momentum ($p$) is essential for relating energy dissipation to collision dynamics. By substituting $v = p/m$ into the definition of kinetic energy:
The Law of Conservation of Linear Momentum states that for an isolated system (where the net external force is zero), the total linear momentum of the system $\vec{P}_{total}$ remains constant.
Proof for a General System of Particles:
The total net force on a system is the sum of all external forces $\vec{F}_{ext}$ and all internal forces $\vec{F}_{int}$:
By Newton's Third Law, the internal forces cancel in pairs ($\sum \vec{F}_{int} = 0$).
If the system is isolated, the net external force is zero ($\sum \vec{F}_{ext} = 0$). This proves the conservation:
Thus, the total initial momentum ($\vec{P}_i$) equals the total final momentum ($\vec{P}_f$): $\vec{P}_{i} = \vec{P}_{f}$.
In all collisions between objects in an isolated system, momentum is conserved. Collisions are classified based on whether kinetic energy ($K$) is conserved.
In a perfectly inelastic collision, the maximum possible amount of kinetic energy is lost, and the colliding objects stick together after the impact, moving with a common final velocity $\vec{v}_{f}$.
Momentum Conservation (1D):
An elastic collision is one where both momentum and total kinetic energy are conserved.
Governing Equations (1D):
1. Conservation of Momentum:
2. Conservation of Kinetic Energy:
The algebraic combination of equations (10) and (11) results in the relative velocity relationship, which is often the most practical starting point for elastic collision problems:
This equation shows that the relative speed of approach equals the relative speed of separation.
The coefficient of restitution ($e$) is defined as the negative ratio of the relative velocity after the collision to the relative velocity before the collision, providing a measure of elasticity:
The Center of Mass (CM) $\vec{R}_{CM}$ is the mass-weighted average position of the system's particles.
The total linear momentum $\vec{P}_{total}$ is simply the total mass $M$ multiplied by the velocity of the center of mass $\vec{V}_{CM}$:
The acceleration of the CM is determined only by the net external force acting on the system. Taking the time derivative of the total momentum (Equation 15):
Combining this with Newton's Second Law for a system (Equation 7, where $\sum \vec{F}_{ext} = d\vec{P}_{total}/dt$):
This proves that the Center of Mass behaves as if the total mass were concentrated at that point, acted upon only by external forces. During collisions and explosions in an isolated system, $\sum \vec{F}_{ext} = 0$, meaning $\vec{A}_{CM} = 0$, and thus $\vec{V}_{CM}$ is constant.
This set includes 5 Intermediate, 3 Advanced, and 2 challenging Irodov-like problems requiring multi-step analysis across different physical principles.
Impulse is the change in momentum: $J = \Delta p = m(v_f - v_i)$. Let $v_i = +40 \text{ m/s}$ (initial direction) and $v_f = -50 \text{ m/s}$ (opposite direction).
$$J = (0.145 \text{ kg})(-50 \text{ m/s} - 40 \text{ m/s}) = (0.145 \text{ kg})(-90 \text{ m/s}) = -13.05 \text{ N \cdot s}$$ $$\text{The magnitude of the impulse is } 13.05 \text{ N \cdot s}$$Using Conservation of Momentum for a perfectly inelastic collision:
$$v_{f} = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1 + m_2}$$ $$v_{f} = \frac{(1500 \text{ kg})(20 \text{ m/s}) + (1000 \text{ kg})(0)}{2500 \text{ kg}}$$ $$v_{f} = \frac{30000}{2500} = 12 \text{ m/s}$$Total initial momentum is zero. $m_{astro} v_{astro} + m_{cam} v_{cam} = 0$.
$$v_{astro} = - \frac{m_{cam} v_{cam}}{m_{astro}} = - \frac{(0.5 \text{ kg})(10 \text{ m/s})}{75 \text{ kg}}$$ $$v_{astro} \approx -0.0667 \text{ m/s}$$ $$\text{The recoil speed is } 0.0667 \text{ m/s}$$From P1, the magnitude of the impulse is $J = 13.05 \text{ N \cdot s}$.
$$\text{Average Force: } F_{avg} = \frac{J}{\Delta t} = \frac{13.05 \text{ N \cdot s}}{0.0035 \text{ s}}$$ $$F_{avg} \approx 3728.6 \text{ N}$$Total Mass: $M = 3 + 4 + 5 = 12 \text{ kg}$.
$$x_{CM} = \frac{(3)(2) + (4)(0) + (5)(-1)}{12} = \frac{1}{12} \text{ m}$$ $$y_{CM} = \frac{(3)(0) + (4)(3) + (5)(-1)}{12} = \frac{7}{12} \text{ m}$$ $$\text{Center of Mass Position: } \left(\frac{1}{12} \text{ m}, \frac{7}{12} \text{ m}\right)$$For an elastic collision where $v_{2i}=0$, the final velocity of $m_1$ is:
$$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i}$$ $$-2 \text{ m/s} = \frac{m_1 - m_2}{m_1 + m_2} (10 \text{ m/s})$$ $$-\frac{1}{5} = \frac{m_1 - m_2}{m_1 + m_2} \implies -(m_1 + m_2) = 5(m_1 - m_2)$$ $$-m_1 - m_2 = 5m_1 - 5m_2 \implies 4m_2 = 6m_1$$ $$\text{The ratio } m_1/m_2 \text{ is } \frac{2}{3}$$Total momentum is conserved: $\vec{P}_i = 0$. Thus, $\vec{p}_C = -(\vec{p}_A + \vec{p}_B)$.
$$\vec{p}_A = (5 \text{ kg})(80 \hat{y}) = 400 \hat{y} \text{ N \cdot s}$$ $$\vec{p}_B = (10 \text{ kg})(40 \hat{x}) = 400 \hat{x} \text{ N \cdot s}$$ $$\vec{p}_C = -(400 \hat{x} + 400 \hat{y}) \text{ N \cdot s}$$ $$\text{Magnitude } p_C = \sqrt{400^2 + 400^2} = 400\sqrt{2} \text{ N \cdot s}$$ $$\text{Speed } v_C = \frac{p_C}{m_C} = \frac{400\sqrt{2} \text{ N \cdot s}}{5 \text{ kg}} = 80\sqrt{2} \text{ m/s}$$ $$\text{Speed of C is approximately } 113.1 \text{ m/s}$$Velocity before bounce ($v_i$): $m g H = \frac{1}{2} m v_i^2 \implies v_i = \sqrt{2 g H}$.
Velocity after bounce ($v_f$): $\frac{1}{2} m v_f^2 = m g h \implies v_f = \sqrt{2 g h}$.
Coefficient of Restitution (since floor is stationary, $v_{2i}=v_{2f}=0$):
$$e = - \frac{v_{f} - 0}{v_{i} - 0} = \frac{|v_f|}{|v_i|} = \frac{\sqrt{2 g h}}{\sqrt{2 g H}}$$ $$e = \sqrt{\frac{h}{H}}$$a) Final velocity $v_{2f}$: Use Conservation of Momentum and the definition of $e$ (Equation 13 with $v_{2i}=0$):
$$\text{Momentum: } m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f} \implies v_{1f} = \frac{m_1 v_{1i} - m_2 v_{2f}}{m_1}$$ $$\text{Restitution: } e = - \frac{v_{1f} - v_{2f}}{v_{1i}} \implies v_{1f} = v_{2f} - e v_{1i}$$ $$\text{Equating } v_{1f} \text{ expressions: } \frac{m_1 v_{1i} - m_2 v_{2f}}{m_1} = v_{2f} - e v_{1i}$$ $$m_1 v_{1i} - m_2 v_{2f} = m_1 v_{2f} - m_1 e v_{1i}$$ $$v_{2f} (m_1 + m_2) = v_{1i} (m_1 + m_1 e)$$ $$v_{2f} = \frac{m_1 v_{1i} (1+e)}{m_1 + m_2}$$b) Kinetic Energy Lost $\Delta K$: $K_{lost} = K_i - K_f = \frac{1}{2} m_1 v_{1i}^2 - \left( \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \right)$.
$$\text{Using the CM velocity } V_{CM} = \frac{m_1 v_{1i}}{m_1 + m_2}.$$ $$\text{Kinetic energy in the CM frame } K'_{i} = \frac{1}{2} \mu v_{rel}^2 \text{ where } \mu = \frac{m_1 m_2}{m_1 + m_2} \text{ and } v_{rel}=v_{1i}.$$ $$\text{The kinetic energy conserved in the collision is the CM kinetic energy: } K_{CM} = \frac{1}{2} (m_1+m_2) V_{CM}^2.$$ $$\text{The energy lost is the loss of kinetic energy relative to the CM:}$$ $$|\Delta K| = K'_{i} - K'_{f} = K'_{i} (1 - e^2) = \frac{1}{2} \mu v_{1i}^2 (1 - e^2)$$ $$\text{Initial Kinetic Energy } K_i = \frac{1}{2} m_1 v_{1i}^2$$ $$\frac{|\Delta K|}{K_i} = \frac{\frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} v_{1i}^2 (1 - e^2)}{\frac{1}{2} m_1 v_{1i}^2} = \frac{\frac{m_1 m_2}{m_1 + m_2}}{m_1} (1 - e^2)$$ $$\frac{|\Delta K|}{K_i} = \frac{m_2}{m_1 + m_2} (1 - e^2)$$This is a three-stage problem: 1. Slide down incline (Energy Conserved), 2. Collision (Momentum Conserved), 3. Slide on rough surface (Work-Energy Theorem).
$$\text{Stage 1: Velocity } v_1 \text{ at bottom of incline (Energy Conservation)}$$ $$v_1 = \sqrt{2 g h} = \sqrt{2 (9.8 \text{ m/s}^2)(5 \text{ m})} \approx 9.90 \text{ m/s}$$ $$\text{Stage 2: Velocity } v_f \text{ after inelastic collision (Momentum Conservation)}$$ $$v_f = \frac{m_1 v_1}{m_1 + m_2} = \frac{(4 \text{ kg})(9.90 \text{ m/s})}{10 \text{ kg}} = 3.96 \text{ m/s}$$ $$\text{Stage 3: Distance } d \text{ on rough surface (Work-Energy Theorem)}$$ $$\text{Work done by friction } W_f = \Delta K$$ $$- F_k d = K_f - K_i \implies - \mu_k M g d = 0 - \frac{1}{2} M v_f^2$$ $$d = \frac{v_f^2}{2 \mu_k g} = \frac{(3.96 \text{ m/s})^2}{2 (0.2) (9.8 \text{ m/s}^2)} = \frac{15.6816}{3.92} \approx 4.00 \text{ m}$$ $$\text{The combined mass travels approximately } 4.00 \text{ m}$$