Linear momentum is a vector quantity defined as the product of an object's mass and its velocity. It serves as a measure of the quantity of motion contained in a translational body.
Definition, units, dimensions, vector nature, conservation of linear momentum in one and two dimensions, application to recoil of guns, collisions in a straight line (elastic and inelastic).
Generalization to systems of many particles, relationship with Newton's second law in differential and integral forms, impulse-momentum theorem, variable mass systems (rocket propulsion).
Conditions: Force acting over a time interval t_1 to t_2
Result: Delta p = J
Conditions: Variable mass system moving in free space with no external forces
Result: v - $v_0 = v_rel \ln(m_0 / m)$
| Formula / law | Equation | Variables and units | Conditions |
|---|---|---|---|
| Linear Momentum Vector | $$p = m v$$ | p = linear momentum, m = mass, v = velocity vector kg m s^-1 | Classical mechanics regime (v << c) |
| Force as Rate of Change of Momentum | $$F = \frac{dp}{dt}$$ | F = net external force, p = linear momentum, t = time N | Inertial frame of reference |
| Linear Impulse | $$J = \int_{t_1}^{t_2} F dt = \Delta p$$ | J = impulse, F = force, Delta p = change in momentum N s or kg m s^-1 | General time-varying or constant force |
| Conservation of Momentum (Two Bodies) | $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$ | u_1, $u_2 = initial velocities; v_1$, $v_2 = final velocities$ kg m s^-1 | Isolated system, no external force |
| Relation Between Kinetic Energy and Momentum | $$E_k = \frac{p^2}{2m}$$ | $E_k = kinetic energy$, p = magnitude of momentum, m = mass J | Non-relativistic particles |
| Velocity of Centre of Mass | $$V_{cm} = \frac{\sum m_i v_i}{\sum m_i}$$ | $V_{cm} = velocity of centre of mass$, m_i and $v_i = mass$ and velocity of i-th particle m s^-1 | Multi-particle system |
| Tsiolkovsky Rocket Equation | $$\Delta v = v_{rel} \ln\left(\frac{m_0}{m}\right)$$ | Delta v = change in speed, $v_rel = relative exhaust velocity$, $m_0 = initial mass$, m = final mass m s^-1 | Motion in free space (zero external force) |
| Discharge Rate in Variable Mass Systems | $$F_{thrust} = v_{rel} \left(-\frac{dm}{dt}\right)$$ | $F_thrust = thrust force$, $v_rel = relative velocity$, dm/dt = rate of mass change N | Steady mass ejection |
Draw: A curve showing force on the vertical axis and time on the horizontal axis.
Interpret: The area enclosed between the curve and the time axis represents the impulse delivered to the body during that time interval.
Exam use: Numerical and conceptual questions asking to calculate change in momentum from graphical areas.
Draw: A straight line passing through the origin for a constant mass particle.
Interpret: The slope of the straight line graph gives the inertial mass ($m = p/v) of the object$.
Exam use: Identifying physical properties from experimental slope interpretations.
Draw: A schematic of a bullet striking and embedding into a suspended block, causing a vertical rise h.
Interpret: Combines the conservation of linear momentum during the collision with the conservation of mechanical energy during the subsequent swing.
Exam use: Standard apparatus-based problem testing simultaneous application of momentum and energy conservation.
Answer: 0 m/s
Answer: Thrust = 20000 N, Acceleratio$n = 40 m/s^2$
Answer: 3v/2 in the original direction of motion
Common error: Kinetic energy is always conserved in any collision where linear momentum is conserved.
Correct: Linear momentum is conserved in all collisions in the absence of external forces, whereas kinetic energy is only conserved in perfectly elastic collisions.
In inelastic collisions, mechanical energy is dissipated into heat, sound, or deformation work while momentum remains conserved.
Common error: If the net external force on a system is zero, individual particles within the system cannot change their momentum.
Correct: Individual particles can experience large changes in internal momentum due to internal forces.
Conservation applies only to the vector sum of momenta of all particles, not to individual components.
| Parameter | Linear Momentum | Kinetic Energy |
|---|---|---|
| Nature | Vector quantity | Scalar quantity |
| Dependence on velocity | Proportional to $v$ | Proportional to $v^2$ |
| Conservation | Conserved in all isolated collisions | Conserved only in elastic collisions |
Choose an option to reveal the explanation.
Explanation: Linear momentum is given by $p = mv$. Mass has dimensions $[M]$ and velocity has dimensions $[L T^{-1}]$, yielding $[M L T^{-1}]$.
Explanation: Newton-second is $\text{N s} = (kg \cdot m \cdot s^{-2}) \cdot s = \text{kg m s}^{-1}$, which matches the unit of momentum.
Explanation: Newton's second law in its general form states that $\vec{F}_{ext} = \frac{d\vec{p}}{dt}$.
Explanation: Impulse is defined as $J = \int F dt$, which geometrically corresponds to the area under the force-versus-time curve.
Explanation: The conservation of linear momentum requires that the net external force acting on the system is zero.
Explanation: From $p = mv$ and $E_k = \frac{1}{2}mv^2$, substituting $v = \frac{p}{m}$ gives $E_k = \frac{p^2}{2m}$.
Explanation: Since $E_k \propto p^2$, doubling $p$ results in $(2)^2 = 4$ times the original kinetic energy.
Explanation: In all collisions without external forces, linear momentum is conserved. Kinetic energy is not conserved in inelastic collisions.
Explanation: The thrust force on a variable mass system is given by $F_{thrust} = v_{rel} \left(-\frac{dm}{dt}\right)$, where $-\frac{dm}{dt}$ is the rate of mass ejection.
Explanation: The basic Tsiolkovsky equation assumes motion in free space where external forces like gravity and drag are neglected.
Explanation: By Newton's third law, internal forces occur in equal and opposite pairs, summing to zero and thus not affecting the total system momentum.
Explanation: Internal explosive forces do not affect the motion of the center of mass; only external forces (like gravity) alter its trajectory.
Explanation: The velocity of the center of mass is the mass-weighted average of the velocities of the constituent particles: $V_{cm} = \frac{\sum m_i v_i}{\sum m_i}$.
Explanation: Momentum is always conserved in isolated systems regardless of collision type, whereas kinetic energy is only conserved in elastic collisions.
Explanation: Since $\vec{F} = \frac{d\vec{p}}{dt}$, integrating both sides over time yields $\int \vec{F} dt = \Delta \vec{p}$, which is the impulse-momentum theorem derived from Newton's second law.
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
Explanation: By conservation of linear momentum: $mv + 0 = (m + M)v'$, which gives $v' = \frac{mv}{m + M}$.
Explanation: Impulse $\Delta p = \int_{0}^{4} (2t + 3) dt = \left[ t^2 + 3t \right]_0^4 = (16 + 12) = 28 \text{ N s}$. Wait, $16+12=28$. Let's check options: 28 is option B.
Explanation: Since $E_k = \frac{p^2}{2m}$, $p = \sqrt{2m E_k}$. For equal $E_k$, $\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}$.
Explanation: Thrust force $F_{thrust} = v_{rel} \mu$. By Newton's second law, acceleration $a = \frac{F_{thrust}}{m} = \frac{\mu v_{rel}}{m}$.
Explanation: Impulse $\Delta p = m(v_f - v_i) = 0.5 \times (-10 - 10) = -10 \text{ N s}$. Magnitude is $10 \text{ N s}$.
Explanation: By momentum conservation: $0 = mv + 2m v_2 \implies v_2 = -v/2$. Kinetic energy = $\frac{1}{2}m v^2 + \frac{1}{2}(2m)(-v/2)^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$. Wait: check $2m(-v/2)^2 = 2m(v^2/4) = \frac{1}{2}mv^2$. Total KE = $\frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
Attempt these questions under a ten-minute limit using a separate timer.
Explanation: Impulse equals change in momentum, so it shares the same dimensions: $[M L T^{-1}]$. mols/units match momentum.
Explanation: In the absence of external forces, the total linear momentum vector of the system is strictly conserved.
Explanation: Since $E_k = \frac{p^2}{2m}$, for equal $p$, $E_k \propto \frac{1}{m}$. Thus $E_{k1} / E_{k2} = m_2 / m_1 = 3 / 1$.
Explanation: From Tsiolkovsky equation, $\Delta v = v_{rel} \ln(m_0 / (m_0/e)) = v_{rel} \ln(e) = v_{rel}$. (Since $\ln(e) = 1$).
Explanation: Total kinetic energy can change via internal processes (e.g., explosions, chemical reactions, springs) even when net external force and total momentum are zero.
Explanation: Change in momentum equals impulse: $\Delta p = F \cdot \Delta t = 5 \text{ N} \times 3 \text{ s} = 15 \text{ N s}$. Since initial momentum is zero, final momentum is $15 \text{ N s}$.
Explanation: The recoil of a gun is a classic application of the conservation of linear momentum in an isolated system.
Explanation: By Newton's second law for a system of particles, $\frac{d\vec{P}}{dt} = \vec{F}_{ext}$ because internal forces cancel out in pairs.
Explanation: Using $E_k = \frac{p^2}{2m}$, $E_k = \frac{6^2}{2(2)} = \frac{36}{4} = 9 \text{ J}$.
Explanation: Since the system is isolated and initially at rest, the total linear momentum remains zero throughout the process.