Concept Card: Linear Momentum & Newton’s Second Law of Motion
1. Linear Momentum ($\vec{p}$):
The linear momentum $\vec{p}$ of a body of mass $m$ moving with velocity $\vec{v}$ is defined as:
$\vec{p} = m\vec{v}$
– Vector quantity; directed along the velocity vector $\vec{v}$.
– SI unit: $\text{kg}\cdot\text{m/s}$ or $\text{N}\cdot\text{s}$. Dimensions: $[M L T^{-1}]$.
Relation with Kinetic Energy ($K$):
$K = \frac{p^2}{2m} \iff p = \sqrt{2mK}$
– For constant kinetic energy: $p \propto \sqrt{m}$ (a heavier body has greater momentum than a lighter body having identical kinetic energy).
– For constant linear momentum: $K \propto \frac{1}{m}$ (a lighter body has greater kinetic energy than a heavier body having identical momentum).
2. Newton’s Second Law of Motion:
Formal Statement: The time rate of change of linear momentum of a body is directly proportional to the net applied external force and takes place in the direction of the applied force.
Fundamental formulation:
$$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$$
For a system of constant mass ($m = \text{constant}$):
$\vec{F}_{\text{net}} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}$
Component equations:
$F_x = \frac{dp_x}{dt} = ma_x, \quad F_y = \frac{dp_y}{dt} = ma_y, \quad F_z = \frac{dp_z}{dt} = ma_z$
Crucial Vector Principle: Force applied along one Cartesian direction affects acceleration exclusively along that axis.
3. Impulse of a Force ($\vec{J}$) & Impulse-Momentum Theorem:
Impulse is defined as the time-integral of force acting over a duration $[t_1, t_2]$:
$\vec{J} = \int_{t_1}^{t_2} \vec{F}\,dt$
Impulse-Momentum Theorem:
$$\vec{J} = \int_{t_1}^{t_2} \frac{d\vec{p}}{dt}\,dt = \vec{p}_2 – \vec{p}_1 = \Delta\vec{p}$$
- Geometric Property: The area under the force-time ($F-t$) curve gives the net impulse $\vec{J}$ and total momentum change $\Delta\vec{p}$.
- Average Force: $\vec{F}_{\text{avg}} = \frac{\vec{J}}{\Delta t} = \frac{\Delta\vec{p}}{\Delta t}$. Increasing the collision contact time $\Delta t$ significantly diminishes the destructive peak force $\vec{F}_{\text{avg}}$ (e.g., catching a cricket ball by drawing hands backward, automotive airbags, shock absorbers).
- Rebound vs. Sticking: An elastic rebound ($\vec{v} \to -\vec{v}$) causes $\Delta p = 2mv$, imparting double the impulse of an inelastic stop ($\Delta p = mv$).
4. Variable Mass Dynamics:
- Conveyor Belt with Falling Sand: If mass is dropped at rate $\frac{dm}{dt}$ onto a belt moving at constant speed $v$:
Driving force required: $F_{\text{ext}} = v\frac{dm}{dt}$
Power supplied by motor: $P = F_{\text{ext}} v = v^2\frac{dm}{dt}$
Rate of increase of sand’s kinetic energy: $\frac{dK}{dt} = \frac{1}{2}\left(\frac{dm}{dt}\right)v^2$
Energy Trap: Exactly $50\%$ of the motor’s power is converted into kinetic energy; the remaining $50\%$ is dissipated as heat during sliding friction before sand catches up to belt speed. - Rocket Propulsion: Thrust force $F_{\text{thrust}} = u_{\text{rel}}\left(-\frac{dm}{dt}\right)$.
Net upward acceleration in gravity: $a = \frac{u_{\text{rel}}}{m}\left(-\frac{dm}{dt}\right) – g$.
Solved Examples
Example 1 (Calculus Differentiation & Impulse-Momentum Link):
A particle of mass $m = 0.50\text{ kg}$ moves in the $x-y$ plane with position vector:
$\vec{r}(t) = (3t^2 – 2t)\hat{i} + (4t^3)\hat{j}\text{ meters}$
(a) Find the linear momentum $\vec{p}(t)$ of the particle.
(b) Find the net force $\vec{F}(t)$ acting on the particle at $t = 2.0\text{ s}$.
(c) Calculate the impulse delivered to the particle from $t = 0$ to $t = 2.0\text{ s}$.
Solution:
(a) $\vec{v}(t) = \frac{d\vec{r}}{dt} = (6t – 2)\hat{i} + (12t^2)\hat{j}\text{ m/s}$.
$\vec{p}(t) = m\vec{v}(t) = 0.5[(6t – 2)\hat{i} + (12t^2)\hat{j}] = (3t – 1)\hat{i} + (6t^2)\hat{j}\text{ kg}\cdot\text{m/s}$.
(b) $\vec{F}(t) = \frac{d\vec{p}}{dt} = 3\hat{i} + 12t\hat{j}\text{ N}$.
At $t = 2.0\text{ s}$: $\vec{F}(2) = 3\hat{i} + 24\hat{j}\text{ N}$. Magnitude $F = \sqrt{3^2 + 24^2} = \sqrt{585} \approx 24.19\text{ N}$.
(c) By Impulse-Momentum Theorem:
$\vec{p}(0) = -1\hat{i}\text{ kg}\cdot\text{m/s}, \quad \vec{p}(2) = 5\hat{i} + 24\hat{j}\text{ kg}\cdot\text{m/s}$.
$\vec{J} = \vec{p}(2) – \vec{p}(0) = (5\hat{i} + 24\hat{j}) – (-1\hat{i}) = 6\hat{i} + 24\hat{j}\text{ N}\cdot\text{s}$.
Magnitude: $J = \sqrt{6^2 + 24^2} = 6\sqrt{17}\text{ N}\cdot\text{s} \approx 24.74\text{ N}\cdot\text{s}$.
Example 2 (Water Jet Wall Impact & Machine Gun Recoil):
(a) A stream of water issuing horizontally with speed $v = 15\text{ m/s}$ from a pipe of cross-sectional area $A = 10^{-2}\text{ m}^2$ strikes a vertical wall normally. The density of water is $\rho = 1000\text{ kg/m}^3$. Assuming water splashes parallel to the wall without rebounding, calculate the force exerted on the wall.
(b) What would the force be if the water rebounded elastically with the same speed?
(c) A machine gun fires $n = 300\text{ bullets}$ per minute, each of mass $m = 20\text{ g}$ with muzzle speed $u = 600\text{ m/s}$. Find the average force required to keep the gun stationary.
Solution:
(a) Mass of water per second: $\frac{dm}{dt} = \rho A v = (1000)(10^{-2})(15) = 150\text{ kg/s}$.
Force on wall (no rebound): $F = \frac{dp}{dt} = \rho A v^2 = (150)(15) = 2250\text{ N}$.
(b) For elastic rebound ($\Delta v = 2v$): $F_{\text{elastic}} = 2\rho A v^2 = 2(2250) = 4500\text{ N}$.
(c) Bullet frequency: $f = \frac{300}{60} = 5\text{ bullets/s}$.
$F_{\text{avg}} = f \cdot (mu) = 5 \times (0.020 \times 600) = 5 \times 12 = 60\text{ N}$.
Example 3 ($F-t$ Trapezoidal Graph & Work-Energy Application):
A variable force $F(t)$ acts on a body of mass $m = 2.0\text{ kg}$ initially at rest. The force increases linearly from $0$ to $20\text{ N}$ in $2.0\text{ s}$, stays constant at $20\text{ N}$ for $4.0\text{ s}$ (from $t = 2\text{ s}$ to $6\text{ s}$), and drops linearly to $0$ in $2.0\text{ s}$ (from $t = 6\text{ s}$ to $8\text{ s}$).
(a) Calculate the total impulse delivered from $t = 0$ to $8.0\text{ s}$.
(b) Find the final speed of the body at $t = 8.0\text{ s}$.
(c) Find the total work done on the body by this force.
Solution:
(a) $J = \text{Trapezoid Area} = \frac{1}{2}(b_1 + b_2)h = \frac{1}{2}(8.0 + 4.0)(20) = \frac{1}{2}(12.0)(20) = 120\text{ N}\cdot\text{s}$.
(b) $J = m(v_f – v_i) \implies 120 = 2.0 v_f \implies v_f = 60\text{ m/s}$.
(c) By Work-Energy Theorem: $W = \Delta K = \frac{1}{2}m v_f^2 – 0 = \frac{1}{2}(2.0)(60^2) = 3600\text{ Joules}$.
Example 4 (Variable Mass: Conveyor Belt & Rocket Thrust):
(a) Sand is dropped vertically at a steady rate $\frac{dm}{dt} = 4.0\text{ kg/s}$ onto a horizontal conveyor belt moving at constant speed $v = 3.0\text{ m/s}$. Find the external force and motor power required, and show that power delivered is twice the rate of kinetic energy gain.
(b) A rocket of initial mass $M_0 = 800\text{ kg}$ expels exhaust gases at relative speed $u_{\text{rel}} = 1200\text{ m/s}$. Taking $g = 10\text{ m/s}^2$, find the rate of fuel consumption required to give the rocket an initial upward acceleration of $a = 20\text{ m/s}^2$.
Solution:
(a) $F_{\text{ext}} = v\frac{dm}{dt} = (3.0)(4.0) = 12.0\text{ N}$.
Motor power: $P = F_{\text{ext}} v = v^2\frac{dm}{dt} = (3.0^2)(4.0) = 36.0\text{ W}$.
$\frac{dK}{dt} = \frac{1}{2}\left(\frac{dm}{dt}\right)v^2 = \frac{1}{2}(4.0)(9.0) = 18.0\text{ W}$.
$P = 2\frac{dK}{dt}$. The remaining $18.0\text{ W}$ is dissipated as heat due to sliding friction before sand reaches belt speed.
(b) $M_0 a = u_{\text{rel}}\left(-\frac{dm}{dt}\right) – M_0 g \implies u_{\text{rel}}\left(-\frac{dm}{dt}\right) = M_0(a + g) = (800)(20 + 10) = 24000\text{ N}$.
Rate of fuel consumption: $-\frac{dm}{dt} = \frac{24000}{1200} = 20\text{ kg/s}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
If the linear momentum of a particle is increased by $50\%$, the percentage increase in its kinetic energy is:
(A) $50\%$
(B) $100\%$
(C) $125\%$
(D) $225\%$
Problem 2 (JEE Main – Single Correct):
A force $\vec{F} = (6t)\hat{i}\text{ N}$ acts on a particle of mass $2\text{ kg}$ initially at rest from $t = 0$ to $t = 3\text{ s}$. The velocity of the particle at $t = 3\text{ s}$ is:
(A) $9\text{ m/s}$
(B) $13.5\text{ m/s}$
(C) $18\text{ m/s}$
(D) $27\text{ m/s}$
Problem 3 (JEE Main – Single Correct):
A rubber ball of mass $m$ strikes a rigid vertical wall with speed $v$ at an angle of incidence $\theta$ to the normal and reflects elastically with the same speed at angle $\theta$. The magnitude of impulse imparted to the ball by the wall is:
(A) $2mv\sin\theta$
(B) $2mv\cos\theta$
(C) $mv\cos\theta$
(D) Zero
Problem 4 (JEE Main – Single Correct):
A machine gun of mass $10\text{ kg}$ fires $10\text{ bullets}$ per second, each of mass $25\text{ g}$ with a speed of $400\text{ m/s}$. The force needed to hold the gun in position is:
(A) $50\text{ N}$
(B) $100\text{ N}$
(C) $200\text{ N}$
(D) $1000\text{ N}$
Problem 5 (JEE Main – Single Correct):
A cricket player catches a ball of mass $0.15\text{ kg}$ moving at $20\text{ m/s}$ and brings it to rest in $0.1\text{ s}$. The average force exerted by the player’s hands on the ball is:
(A) $15\text{ N}$
(B) $30\text{ N}$
(C) $150\text{ N}$
(D) $300\text{ N}$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are correct regarding Newton’s Second Law of Motion?
(A) The equation $\vec{F} = m\vec{a}$ is valid strictly only when mass $m$ is constant.
(B) Force is a vector quantity that represents the instantaneous rate of momentum transfer.
(C) If the net force on a particle is zero, its linear momentum must remain constant in magnitude and direction.
(D) The rate of change of kinetic energy of a particle equals the instantaneous power: $\frac{dK}{dt} = \vec{F}\cdot\vec{v}$.
Problem 7 (JEE Advanced – One or More Correct):
A particle of mass $m$ collides elastically with a smooth stationary vertical wall. Let the normal to the wall be along the $x$-axis and the wall surface lie in the $y-z$ plane. Which of the following is/are correct?
(A) The component of linear momentum parallel to the wall ($p_y, p_z$) is conserved.
(B) The component of linear momentum perpendicular to the wall ($p_x$) reverses sign ($p_{x,f} = -p_{x,i}$).
(C) The impulse imparted by the wall on the particle is directed purely along the normal to the wall.
(D) The kinetic energy of the particle is conserved.
Problem 8 (JEE Advanced – One or More Correct):
Sand is continuously dumped at rate $\mu = \frac{dm}{dt}$ onto a flat horizontal conveyor belt driven by an electric motor at constant speed $v$. Which of the following statements is/are correct?
(A) The external driving force required to maintain the belt’s speed is $\mu v$.
(B) The power delivered by the motor is $\mu v^2$.
(C) The rate of increase of kinetic energy of the sand is $\frac{1}{2}\mu v^2$.
(D) The rate of heat dissipation due to friction between sand and belt is $\frac{1}{2}\mu v^2$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A baseball of mass $m = 0.15\text{ kg}$ traveling horizontally at $v_1 = 30\text{ m/s}$ is hit by a bat and flies straight back at $v_2 = 40\text{ m/s}$. If the duration of contact is $\Delta t = 0.005\text{ s}$ ($5\text{ ms}$), the magnitude of the average force exerted on the ball by the bat in Newtons is $F_{\text{avg}}$. Find $F_{\text{avg}}$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A rocket of initial mass $M = 1200\text{ kg}$ is fired vertically upward. The relative exhaust speed of the ejected gases is $u_{\text{rel}} = 600\text{ m/s}$. Taking $g = 10\text{ m/s}^2$, find the rate of fuel burning in $\text{kg/s}$ required to give the rocket an initial net upward acceleration of $15\text{ m/s}^2$.
Solutions & Explanations
Answer Key Summary:
1. (C) | 2. (B) | 3. (B) | 4. (B) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 2100 | 10. 50
Solution 1:
$K_1 = \frac{p^2}{2m}$. If $p_2 = 1.5p$, $K_2 = \frac{(1.5p)^2}{2m} = 2.25 K_1$. Percentage increase $= \frac{2.25K_1 – K_1}{K_1} \times 100\% = 125\%$.
Correct Answer: (C)
Solution 2:
$F = m\frac{dv}{dt} \implies 6t = 2\frac{dv}{dt} \implies \frac{dv}{dt} = 3t \implies v = \int_{0}^{3} 3t\,dt = \left[\frac{3t^2}{2}\right]_{0}^{3} = 13.5\text{ m/s}$.
Correct Answer: (B)
Solution 3:
Initial normal momentum $p_{x,i} = mv\cos\theta$; final normal momentum $p_{x,f} = -mv\cos\theta$.
$J = |\Delta p_x| = |-mv\cos\theta – mv\cos\theta| = 2mv\cos\theta$.
Correct Answer: (B)
Solution 4:
$F = n \cdot m \cdot v = 10 \times 0.025 \times 400 = 100\text{ N}$.
Correct Answer: (B)
Solution 5:
$\Delta p = m(0 – v) = -0.15 \times 20 = -3.0\text{ kg}\cdot\text{m/s}$.
$F_{\text{avg}} = \frac{|\Delta p|}{\Delta t} = \frac{3.0}{0.1} = 30\text{ N}$.
Correct Answer: (B)
Solution 6:
All statements (A, B, C, D) are correct formulations of Newton’s second law, momentum conservation, and power.
Correct Answer: (A, B, C, D)
Solution 7:
For a smooth wall, no shear force acts parallel to the surface; momentum parallel to the wall is conserved while normal momentum reverses sign. Collision is elastic, conserving kinetic energy.
Correct Answer: (A, B, C, D)
Solution 8:
All statements (A, B, C, D) are fundamental properties of the conveyor belt variable mass problem.
Correct Answer: (A, B, C, D)
Solution 9:
$\Delta p = m(v_2 – (-v_1)) = 0.15(40 + 30) = 0.15 \times 70 = 10.5\text{ kg}\cdot\text{m/s}$.
$F_{\text{avg}} = \frac{10.5}{0.005} = 2100\text{ N}$.
Correct Answer: 2100
Solution 10:
$M a = u_{\text{rel}}\left(-\frac{dm}{dt}\right) – Mg \implies u_{\text{rel}}\left(-\frac{dm}{dt}\right) = M(a + g) = 1200(15 + 10) = 30000\text{ N}$.
$-\frac{dm}{dt} = \frac{30000}{600} = 50\text{ kg/s}$.
Correct Answer: 50