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

2024
Views:

264

Pages:

5
Author:

Jabin F.

Straight line motion under gravity: object drop from height h a = -g for downward motion a = g for upward motion v = u + at s = ut + (1/2)at² h = (1/2)gt² ⇒ t = √(2h/g) Q. A particle of mass m and charge q is thrown with initial velocity v₀î into a uniform magnetic field B(k) at t = 0. Calculate average force on the particle during the time t = 0 to (πm/qB). Q. A ball of mass m is dropped from a tall building with zero initial velocity. In addition to gravity, the ball experiences a damping force of the form -γv, where v is its instantaneous velocity and γ is a constant. Given values m = 10 kg, γ = 10 kg/s, and g = 10 m/s², the distance travelled (in meters) in time t in seconds is Solution: Method (1) Using EOM: Fnet = ma = mg - γv = mdv/dt ⇒ dv/dt = g - (γ/m)v ⇒ ∫dv/(g - (γ/m)v) = ∫dt ⇒ (-m/γ) ln(mg - γv) = t + C ⇒ v = (mg/γ)(1 - e^(-γt/m)) ⇒ dx/dt = (mg/γ)(1 - e^(-γt/m)) ⇒ ∫dx = ∫(mg/γ)(1 - e^(-γt/m)) dt ⇒ x = (mg/γ)(t + (1/γ)e^(-γt/m)) |₀ᵗ = (mg/γ)(t + (1/γ)e^(-γt/m) - 1) Method (2) Using 2nd order differential equation: ma = mg - γv = mg - γ(dx/dt) ⇒ d²x/dt² + (γ/m)(dx/dt) = g Solution of equation: D² + (γ/m)D = 0 ⇒ D(D + γ/m) = 0 ⇒ D = 0, -γ/m Complementary Function (CF): CF = C₁e^(0t) + C₂e^(-γt/m) = C₁ + C₂e^(-γt/m) Particular Integral (PI): f(D) = D + γ/m As f(0) = 0, therefore PI = (1/f(D)) * 10 = 10/(D + γ/m) = 10