University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

350

Pages:

5
Author:

Jabin F.

# Problem Solving using Newton's EOM: → Any question in which force is mentioned, can be solved using Newton's EOM. Approach: Put net force in the above equation and obtain acceleration at some general instant of time, t. → If forces are constant, then acceleration also constant, then use: v = u + at s = ut + (1/2)at² → If forces are not constant (F = f(x) or f(t) or f(v)), then acceleration is variable, then use: a = dv/dt = d²x/dt² ∫a dx = ∫v dv ∫v dt = ∫x dx → If the question is related to distance and time but ax = d²x/dt² is not solvable, then use ax = v(dv/dx) to find v as a function of x. Then use v = dx/dt and solve the 1st order differential equation to find the relation between x and t. → Take the direction of motion as the positive direction. Q: A bullet with initial speed u₀ is fired at a block of wood. The resistive force by wood on the bullet is given by F = -kv², where k is a constant. What is the time taken to stop the bullet inside the wood block? Solution: F = ma = -kv² m(dv/dt) = -kv² dv/v² = -(k/m)dt ∫(1/v²) dv = -(k/m)∫dt -1/v = -(k/m)t + C 1/v = (k/m)t - C v = 1 / ((k/m)t - C) At t = 0, v = u₀: 1/u₀ = -C C = -1/u₀ Therefore: v = 1 / ((k/m)t + 1/u₀) 0 = 1 / ((k/m)t + 1/u₀) (k/m)t + 1/u₀ = ∞ t = ∞ Q: A small raindrop of mass m experiences a viscous drag force F = -bv, proportional to its instantaneous speed 'v'. If it starts from rest at a height h, its speed after a time 't' is: Solution: Fnet = ma = mg - bv mg - bv = mdv/dt mg - bv = m(dv/dt) (mg - bv)/m = dv/dt dt = m dv / (mg - bv) ∫dt = ∫m dv / (mg - bv) t = -(m/b) ln(mg - bv) + C At t = 0, v = 0: 0 = -(m/b) ln(mg) + C C = (m/b) ln(mg) Therefore: t = -(m/b) ln(mg - bv) + (m/b) ln(mg) t = (m/b) ln(mg) - (m/b) ln(mg - bv) t = (m/b) ln(mg / (mg - bv))