AP Physics C: Mechanics:: Systems Momentum

For a DISCRETE system of point masses, the CENTER OF MASS formula, x_cm=(sum of m_i*x_i)/(sum of m_i), is familiar from algebra-based physics. What is the GENERALIZED, calculus-based formula for the CENTER OF MASS of a CONTINUOUS object (like a rod with varying density), using an INTEGRAL instead of a sum?

- **A)** The continuous center-of-mass formula requires DIFFERENTIATING (rather than integrating) the object's mass distribution, rather than the correct integral-based approach
- **B)** This concept has no actual mathematical relationship between the discrete center-of-mass formula and its continuous, integral-based generalization
- **C)** x_cm = INTEGRAL[x dm] / INTEGRAL[dm], where the sum over discrete point masses (m_i) is replaced by an INTEGRAL over infinitesimally small mass elements (dm) distributed continuously throughout the object -- the denominator, INTEGRAL[dm], simply equals the object's TOTAL mass M
- **D)** x_cm = INTEGRAL[x] / INTEGRAL[m] (integrating position alone, without properly weighting by mass elements), rather than the correct x*dm-weighted integral

For a THIN ROD of length L with UNIFORM (constant) LINEAR MASS DENSITY lambda=M/L, lying along the x-axis from x=0 to x=L, how is the MASS ELEMENT dm EXPRESSED in terms of lambda and the position element dx, to set up the center-of-mass integral?

- **A)** This concept has no actual mathematical relationship between linear mass density and expressing a differential mass element for integration
- **B)** dm = lambda*dx -- since linear mass density (lambda) represents mass PER UNIT LENGTH, multiplying by a small length element dx gives the corresponding small mass element dm, which can then be substituted into the center-of-mass integral, x_cm=INTEGRAL[x dm]/M
- **C)** dm = lambda/dx (dividing rather than multiplying density by the length element), rather than the correct multiplicative relationship
- **D)** dm = lambda + dx (adding rather than multiplying density and the length element), rather than the correct multiplicative relationship

For a THIN ROD of length L with a NON-UNIFORM linear mass density given by lambda(x) = C*x (density INCREASING linearly along the rod's length, where C is a constant), what is the TOTAL MASS M of the rod, found by INTEGRATING dm=lambda(x)*dx from x=0 to x=L?

- **A)** M = C*L^2 (incorrectly omitting the (1/2) factor that results from properly integrating the linear density function)
- **B)** M = C*L (treating density as if it were constant at its average or endpoint value, rather than properly integrating the varying density function)
- **C)** M = INTEGRAL from 0 to L of C*x dx = [(C/2)*x^2] from 0 to L = (C/2)*L^2
- **D)** M = INTEGRAL[C*x]dx = C*x^2 evaluated without the correct antiderivative coefficient, giving C*L^2 rather than (C/2)*L^2

For the SAME non-uniformly-dense rod (lambda(x)=C*x, from x=0 to x=L, total mass M=(C/2)*L^2), what is the CENTER OF MASS position, x_cm, found by evaluating x_cm=INTEGRAL[x*dm]/M?

- **A)** x_cm = INTEGRAL from 0 to L of x*(C*x) dx / M = INTEGRAL from 0 to L of C*x^2 dx / M = [(C/3)*L^3] / [(C/2)*L^2] = (2/3)*L -- notably, this center of mass is CLOSER to the denser end (x=L) than the simple midpoint (L/2) would be
- **B)** x_cm = (1/3)*L, incorrectly placing the center of mass closer to the LESS dense end, contradicting the actual mass-weighted result
- **C)** x_cm = L/2 (the simple geometric midpoint), incorrectly ignoring the non-uniform density distribution entirely
- **D)** This concept has no actual relationship between a non-uniform density function and the resulting center-of-mass position for this specific rod

How is the VELOCITY of a system's CENTER OF MASS, v_cm, related to the DERIVATIVE of the center-of-mass POSITION, x_cm(t), with respect to time, directly paralleling the earlier Kinematics unit's definition of velocity for a single particle?

- **A)** v_cm = dx_cm/dt -- the velocity of a system's center of mass is simply the TIME DERIVATIVE of the center-of-mass POSITION function, exactly paralleling how a single particle's velocity is the derivative of its position (from the earlier Kinematics unit)
- **B)** The center of mass of a system of particles CANNOT actually have a well-defined velocity, unlike a single particle
- **C)** v_cm is calculated as the INTEGRAL (rather than the derivative) of the center-of-mass position function, rather than the correct derivative-based definition
- **D)** This concept has no actual mathematical relationship between a system's center-of-mass velocity and the derivative of its center-of-mass position

What is the TOTAL MOMENTUM of a SYSTEM of particles, expressed in terms of the system's TOTAL MASS and the VELOCITY of its CENTER OF MASS, P_total=M*v_cm, and how is this result derived from the individual momenta of each particle in the system?

- **A)** P_total = M/v_cm (dividing rather than multiplying total mass by center-of-mass velocity), rather than the correct multiplicative relationship
- **B)** P_total has no actual mathematical relationship to the system's total mass or its center-of-mass velocity
- **C)** P_total = M*v_cm can be derived by differentiating the center-of-mass position formula, M*x_cm = sum(m_i*x_i), with respect to time: M*v_cm = sum(m_i*v_i) = sum(p_i) = P_total -- showing that the SUM of individual particle momenta EQUALS the total mass times the center-of-mass velocity
- **D)** This formula applies only to systems with EXACTLY TWO particles, with no actual generalization to systems with more particles

What is IMPULSE, expressed as a DEFINITE INTEGRAL of a TIME-VARYING force, J=INTEGRAL[F(t)]dt, and how does this GENERALIZE the simpler algebra-based formula, J=F*delta_t (valid only for constant force)?

- **A)** J = INTEGRAL from t1 to t2 of F(t) dt -- this definite integral definition correctly calculates impulse for ANY force function, including forces that VARY with time, and REDUCES to the simpler J=F*delta_t formula when force happens to be constant
- **B)** Impulse can only be calculated using integration when force is CONSTANT, with no actual applicability to time-varying forces
- **C)** J = F(t2) - F(t1) (subtracting force VALUES rather than integrating), rather than the correct definite-integral definition
- **D)** This concept has no actual mathematical relationship between impulse and the definite integral of a time-varying force function

What is the mathematical RELATIONSHIP between IMPULSE (as a definite integral of force over time) and the CHANGE IN MOMENTUM of an object, derived by INTEGRATING Newton's second law, F_net(t)=dp/dt, with respect to TIME?

- **A)** This calculus-based derivation only applies to CONSTANT forces, with no actual ability to derive the impulse-momentum theorem for time-varying forces
- **B)** Integrating F_net(t)=dp/dt with respect to time from t1 to t2 gives: INTEGRAL[F_net(t)]dt = INTEGRAL[dp] = p(t2)-p(t1) = delta_p -- exactly showing that IMPULSE (the left side) EQUALS the CHANGE IN MOMENTUM (the right side), a rigorous derivation of the impulse-momentum theorem
- **C)** This integration has no actual relationship to deriving the impulse-momentum theorem from Newton's second law
- **D)** This derivation shows that impulse equals the change in KINETIC ENERGY (rather than momentum), contradicting the actual impulse-momentum theorem