Newton's Second Law and Momentum/Impulse Newton's Second Law and Momentum/Impulse Force changes motion. In everyday life we feel that a gentle push over a longer time can produce the same effect as a sharp hit for a brief moment. Physics wraps this idea in two connected quantities: momentum and impulse. Momentum p =m v measures how hard it is to stop a moving object; impulse J = F ,dt measures the total "push" delivered by a force over an interval. Newton's second law in its most general form says the net external force equals the time rate of change of momentum: F net = d p dt . For constant mass, this reduces to the familiar F net =m a . The impulse–momentum theorem J = p then follows by integrating over time. These are vector relations: directions and signs matter. Once you pick a system (what objects are included), internal forces cancel in pairs and only external forces change the system's total momentum. That is why rockets accelerate not because space "pushes" them, but because high-speed exhaust carries momentum away, letting the rocket's momentum change in the opposite direction. The same ideas help decode ball–bat collisions, airbags and cushioning that extend contact time, and carts that speed up as forces act. This lesson builds your intuition first, then shows the math, boundaries, and classic traps NEET examiners love to set. remember Cushioning saves you: for the same change in momentum, increasing contact time reduces average force (big t → smaller F avg ). Airbags, padded mats, and sand pits use this. Momentum A vector measure of motion: p =m v , units kg ,m/s . Points in the direction of velocity. Impulse Total effect of a force over time: J = t 1 t 2 F ,dt= p , units N ,s = kg ,m/s . The vector sum of all forces on a system due to bodies outside the system. Only this changes the system’s total momentum. Net external force The constant force that would produce the same impulse over a time interval: F avg = p t . Average force Instantaneous force The limit of average force as t 0 : F = d p dt . When m is constant, F =m a . The set of bodies whose total momentum you track. Choose to simplify external forces; internal action–reaction pairs cancel. System Force due to momentum carried away by expelled mass: T = - , m , u rel on the main body (sign by convention). Thrust (variable-mass systems) We will use Cartesian sign convention unless stated: choose a positive direction and keep it fixed. In 1D, vectors reduce to signed numbers; in 2D/3D, work component-wise. Always define your system first: a single ball, ball+bat together, or rocket alone? This choice decides what is "external" and whether momentum is conserved. Valid in inertial frames; p =m v may have time-varying m in special cases like rockets. Newton’s Second Law (General) Constant mass form When m is constant, d p dt =m d v dt =m a . This value represents the total change in momentum ( p ) resulting from a net force applied over a defined time interval. Momentum connects mass and velocity. A truck at low speed can have the same momentum as a light scooter at high speed. Because momentum is linear in velocity, reversing direction flips the sign of p . In vector problems, handle signs carefully: p =m( v f- v i) , not m(| v f|-| v i|) . Momentum SI units kg ,m/s . Vector in direction of v . Calculate the initial or final momentum of an object using this relationship before determining the change in momentum ( p ) due to an impulse. Impulse is the area under the force–time curve. Even if the force spikes rapidly during impact, the integral F ,dt is finite and equals the change in momentum. This is why we can replace a complicated, rapidly varying force with an average force acting over the contact time: they share the same impulse. Impulse (integral form) Total "push" over a time interval. Impulse–Momentum Theorem Directly connects force history to change in momentum. The average force applied over a time interval equals the change in the object's momentum. Graphically, the impulse equals the signed area under F – t . A rectangular pulse of height F 0 lasting t gives J=F 0 , t . A triangular pulse of peak F over base t gives J= 1 2 F , t . If the force reverses sign during the interval, areas above and below the time axis subtract. Time custom Area under each curve equals impulse. Mixed-sign areas subtract. Rectangular pulse from t=a to t=b at +F0 and a triangular pulse from t=c to t=d peaking at +Fmax. Start of rectangular pulse F0 F0 End of rectangular pulse Triangular base start Triangular peak Fmax (c+d)/2 Triangular base end Force F(t) dependent Average force turns impulse into a simple ratio: F avg = p t . In impact problems, always compute the vector change in momentum first, then divide by contact time to get the average force vector. If you also need peak force, you must know the force profile. Same impulse as the actual varying force over t . Average force over an interval General kinematics for any type of motion (constant or non-constant acceleration), provided the time interval is finite and non-zero. Applicability: F = d p dt holds in inertial frames. For constant mass, use F =m a . For variable mass (rockets), do NOT write F =m , d v dt blindly; account for momentum carried by mass flow. tip System thinking: If your system includes all interacting bodies and there is no external force, total momentum is conserved: d P sys dt = 0 . Internal forces are equal and opposite (Newton’s third law) and cancel in the total. If some external forces act, the total momentum changes according to their resultant impulse. Here P sys = i m i v i is the system momentum. System form of Newton’s second law This relationship defines the net external force required to calculate the acceleration of the system's center of mass. Center of mass motion: define total mass M and position R cm . Then M , d 2 R cm dt 2 = F ext . This lets you treat a system as a single particle at its center of mass for translational motion, regardless of internal interactions. Only external forces accelerate the center of mass. Center of mass acceleration In short-contact collisions, forces vary wildly in milliseconds, but the impulse–momentum theorem remains exact. That is why high-speed video and force plates report impulses to characterize hits and landings. In multi-dimensional rebounds (e.g., off a wall), act component-wise: compute p x , p y , then F avg = 1 t ( p x , i + p y , j ) . Use J = F t and J = m (v - u). F = 10 N m = 0.50 kg u = 0 t = 0.20 s Impulse J and final speed v N·s, m/s A constant horizontal force of 10 N acts on a 0.50 kg cart initially at rest for 0.20 s. Find the impulse and the final speed. easy Notice how we first found impulse from force and time, then used p =m( v f- v i) to get v . This flow works even when the force is not constant, as long as you can find the area under the F – t curve. Compute p = m( v f - v i ) with sign; then F avg = p / $ t. m = 0.15 kg v i = +20 m/s ( right ) v f = -30 m/s ( left ) t = 5.0 10 -3 s Average force vector on the ball A 0.15 kg ball moving to the right at 20 m/s is hit by a bat and rebounds to the left at 30 m/s. The contact time is 5.0 ms. Find the magnitude and direction of the average force on the ball. medium Always keep direction signs. Rebound problems typically flip velocity sign. The larger average force magnitude arises because the ball not only stops but also speeds up in the opposite direction, causing a big p over a tiny t . In rocket-alone system, thrust T = m ,u rel (direction opposite exhaust). Then a = T/M. N, m/ s 2 Thrust T and acceleration a of the rocket at that instant M = 100 kg m = 5.0 kg/s ( fuel mass flow rate ) u rel = 400 m/s F ext = 0 A rocket in deep space of instantaneous mass 100 kg ejects fuel at a constant rate 5.0 kg/s with exhaust speed 400 m/s relative to the rocket. Neglect external forces. Find (i) the thrust magnitude and (ii) the rocket’s instantaneous acceleration. hard Sign convention for thrust: if exhaust is expelled backward, thrust on the rocket is forward. In presence of gravity, net force would be T + F ext (e.g., T -Mg j vertically), and a = T + F ext M . Impulse is the same thing as force. Impulse is force accumulated over time. A small force over long time can give the same impulse as a huge force over a short time: J = F ,dt . They cannot. Internal forces cancel in the total, so only external forces change system momentum: F ext = d P sys dt . Internal action–reaction forces can change the total momentum of an isolated system. Vector trap: p =m( v f- v i) , not m(| v f|-| v i|) . For rebounds, one velocity is negative of the other in 1D. neet-alert neet-alert Do not mix up time of flight with contact time. Average force in impacts uses the actual contact duration (typically milliseconds), not the total motion time. Two-dimensional impulse: If a ball hits a smooth vertical wall, the normal component of velocity reverses while the tangential component is unchanged (if no friction). Then p = -2m v , p =0 . Average force points normal to the wall: F avg = p t , n . Define your system and choose a positive direction or axes. Write knowns: m , initial and final velocities, contact time, and any external forces. Compute p =m( v f- v i) component-wise. Use F avg = p t for average force, or integrate F (t) to get impulse. Check units and direction; report with appropriate sign or angle. Workflow for impulse–momentum problems Link to collisions: In an isolated two-body collision (no external impulse in the short interaction), total momentum is conserved though kinetic energy may or may not be. Impulse on each body is equal in magnitude and opposite in direction, consistent with Newton’s third law over the short contact. Angle between the normal and the resultant of normal+limiting friction: = -1 ( s) . Cross-link: helps when converting frictional contact into a single resultant in impact setups. Angle of Friction: = -1 ( s) = -1 ( s) Static friction at limiting value F s= s N Resultant of N and F s makes angle with N Surfaces in contact; impending motion Magnitude of equilibrant (equal to resultant of two forces): F eq = F 1 2+F 2 2+2F 1F 2 . Useful when replacing many contact forces by a single effective force. F eq = F 1 2 + F 2 2 + 2 F 1 F 2 Two forces F 1, F 2 act at a point with angle between them Equilibrant F eq balances F 1+ F 2 in static equilibrium Equilibrant/Resultant magnitude for two concurrent forces Those friction and equilibrant results are cross-topic but appear in momentum/impulse questions where contact forces are combined or their directions need quick estimation. Our main tool remains F = d p dt and its time-integrated form. Quantity Definition Symbol SI Unit Vector? How to compute quickly Force Rate of change of momentum N = kg ,m/s 2 Yes If constant mass: m a ; else slope of p (t) Momentum Mass times velocity kg ,m/s Yes Multiply m and v ; add vectors for systems Impulse Force accumulated over time N ,s Yes Area under F – t curve; J = p Compare the three core ideas Impulse–momentum link J is the Journey under the Force–time curve: J = area under F – t = p . Momentum p : kg ,m/s ; dimension [M][L][T] -1 Impulse J : N ,s = kg ,m/s (same as momentum) Average force: N ; 1 N =1 kg ,m/s 2 Area under F – t graph has units N ,s Units and dimensions custom Slope of p – t is force. Constant slope → constant force. Time Momentum (x-direction) kg·m/s dependent p x (t) derived F x Linear increase of p x (t) indicating constant force; slope equals F x . p0 Initial momentum Final momentum p0 + F x T 2D PLOT x-momentum vs time under a constant force px = p0 + Fx t px p0 Initial momentum Fx Force (x) Another powerful view: in any direction, the slope of the momentum–time graph is the force component, F x= dp x dt . If the graph is steep and positive, you have a large positive force; if it slopes down, the force component is negative. J = p Newton’s second law in general form F net = d p dt Time interval [t 1,t 2] with possibly varying force Impulse–Momentum Theorem J = p In the collision simulator, watch total momentum stay constant (if no external force), while each object’s momentum jumps during the short contact. The jump equals the impulse on that object; the impulses are equal and opposite, matching Newton’s third law integrated over time. Cushions, airbags, crumple zones, and landing mats all work by increasing contact time so that F avg = p/ t becomes smaller for the same p . remember Experimental note: Force plates measure vertical ground reaction during running and jumping. The integral (impulse) predicts take-off velocity. In labs, high-speed force sensors and motion capture cross-check F ,dt with measured m , v to verify J = p within uncertainty. Choosing the right equation set Scenario Mass behavior Correct equation Example Typical pitfall Constant mass m = constant F net = m a Cart pulled by rope Forgetting vector sign Variable mass (control mass approach) m changes via mass flow F ext + T = m a with T = - m , u rel Rocket expelling exhaust Using F =m ,d v /dt without thrust term Isolated system (no external force) Total m may be constant or not if system includes exhaust d P sys dt = 0 Two skaters pushing apart Forgetting to include all bodies in system Boundary checks: If t 0 while p is finite (e.g., a very hard impact), |F avg | in ideal math, but real materials deform to limit peak force. If m 0 at fixed p , velocity becomes huge—unphysical for macroscopic bodies; this flags modeling limits. Closing perspective: Momentum–impulse methods excel when forces are complicated but short-lived, or when internal interactions dominate. Start from the system view, track p with care for directions, and translate between the F – t picture and velocity changes. With these habits, NEET dynamics questions become structured and quick. Key terms recap Linear momentum Momentum Product of mass and velocity, p =m v . Impulsive action Impulse Integral of force over time; equals change in momentum, J = F ,dt= p . F avg = p / t over a specified interval. F avg Average force Net external force Sum of forces on a system from outside; governs d P sys /dt . F ext Thrust Force due to momentum carried away by expelled mass; T = - m , u rel . R cm Weighted average position of mass; moves as if all mass is concentrated there under F ext . Center of mass