Physics · Ch 4 — Laws of Motion

Principle of Conservation of Linear Momentum

4.7

Principle of Conservation of Linear Momentum

From Newton's second law, $\vec{F}=\frac{d\vec{p}}{dt}$ ; if the resultant (net) force is zero, the linear momentum does not change -- it stays constant, i.e. it is CONSERVED. This is the PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM: 'the total momentum of an isolated system is conserved during any interaction.' An isolated system is one with no external force acting on it -- a collection of particles, colliding bodies, or exploding fragments, where the forces of interaction (collision, explosion) are INTERNAL to the system as a whole, even though they may be treated as external forces if each particle is analysed in isolation.

SYSTEMS AND FREE BODY DIAGRAMS: For problems involving several connected masses and pulleys, it is often useful to first identify which masses move identically (and can temporarily be treated as one combined system), and then draw a FREE BODY DIAGRAM (FBD) for each individual body -- a diagram showing ONLY the forces acting on that one body and its own acceleration, nothing about any other body in the system. Consider the illustrative pulley arrangement of Fig 4.2(a): pulleys P1, P3, P4 are fixed while P2 is movable; an applied force F=100 N at 60 degrees to the horizontal acts against a constant opposing (frictional) force of 10 N on a 5 kg mass. As long as the 1 kg mass has not reached pulley P1, the 1 kg and 2 kg masses move identically and can be treated as one 3 kg system (except when the tension $T_3$ between them specifically needs to be found). All the masses except this combined 3 kg system move the same distance in the same time, so their accelerations share the same magnitude; but because P2 is movable, if the upper string moves a distance x, the lower string (holding the 5 kg side) moves only x/2 -- a purely kinematic (pulley) constraint.

For the 2 kg mass (Fig 4.2b), the only two forces are its weight 2g (down) and the tension $T_3$ (up), giving $T_3-2g=2a$ for its upward acceleration a. For the 5 kg mass (Fig 4.2c), the forces are its weight 5g (down), the normal reaction N (up), the opposing frictional force F=10 N (left), and the applied force F=100 N resolved into vertical ( $F\sin60^\circ$ ) and horizontal ( $F\cos60^\circ$ ) components, together with the two string tensions. Resolving vertically gives $N+F\sin60^\circ=5g$ ; resolving horizontally gives the net horizontal driving force. Writing and solving such equations simultaneously for every body in the system yields all the unknown tensions and the common acceleration. …

Figure 4.2aFig 4.2(a): System for illustration of free body diagram

What this figure shows. A composite pulley-and-mass arrangement used to teach the free-body-diagram method: three FIXED pulleys labelled P1, P3 and P4 and one MOVABLE pulley labelled P2. A string carries a 1 kg mass over pulley P1 and continues down to join a 2 kg mass (the string segment between the 1 kg and 2 kg masses is labelled S); together these behave as a single 3 kg system while the 1 kg mass has not yet reached P1. A second string (labelled S1) from the movable pulley P2 supports a 5 kg mass resting against a rough contact surface that offers a constant opposing (frictional) force of F = 10 N; an applied force F = 100 N pulls at an angle of 60 degrees to the horizontal on this 5 kg mass. Tensions in the various string segments are labelled T, T1 and T3 at the relevant junctions/pulleys. Because P2 is movable, if the string S (holding the 1 kg/4 kg masses -- the source text also references a 4 kg mass on the arrangement) moves through a distance x, the lower s …

Figure 4.2bFig 4.2(b): Free body diagram for 2 kg mass

What this figure shows. An isolated free-body sketch of ONLY the 2 kg mass from Fig 4.2(a), drawn as a small block with exactly two force arrows on it: one arrow pointing straight down labelled 2g (its weight, due to Earth), and one arrow pointing straight up labelled T3 (the tension pulling up through the string connecting it to the movable pulley system). No other forces (no normal reaction, no horizontal forces) are shown, since this mass hangs freely. The corresponding equation of motion from Newton's second law is $T_3 - 2g = 2 …

Figure 4.2cFig 4.2(c): Free body diagram for 5 kg mass

What this figure shows. An isolated free-body sketch of ONLY the 5 kg mass resting on the rough horizontal table from Fig 4.2(a). Four/five force arrows act on it: a downward arrow 5g (its weight); an upward arrow N (the normal reaction from the table surface); a leftward arrow F=10 N (the constant opposing/frictional force from the rough contact surface); the applied force F=100 N drawn at 60 degrees above the horizontal (shown resolved into its vertical component F sin60 degrees, adding to the vertical balance N + F sin60 degrees = 5g, and its horizontal component F cos60 degrees); and the two string tensions T and T1 pulling horizontally from the two strings attached on either side of this mass. The mass's ac …

Misc Ex.5Atwood machine -- accelerations and string tension

Worked out. A fixed, frictionless pulley carries a massless inextensible string with masses m1 and m2 (m2 > m1) hanging from its two ends -- the classic Atwood machine. Method I treats the two masses as one system of total mass (m1+m2) driven by the net downward force (m2-m1)g to get a = (m2-m1)g/(m1+m2), then applies Newton's second law to m1 alone (free body diagram: weight m1g down, tension T up, net upward acceleration a) to get T = m1(g+a); Method II instead writes separate free-body equations T - m1g = m1a for the rising mass and m2g - T = m2a for the falling mass and solves the pair simultaneously for the same a and T = 2m1m2g/(m1+m2). A practical tip notes that marking an X on the string …

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