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

2024
Views:

335

Pages:

5
Author:

Jabin F.

# Dependence of area of phase space on energy in 1-D potential phase space: co-ordinates (position & momentum) x, p area of phase space dA = dx dp E = KE + PE = p²/2m + V(x) ⇒ p = √(2m(E - V(x))) ∴ A = ∫√(2m(E - V(x))) dx = E^(1/2 + 1/n) ∫√(2m(1 - V(x)/E)) dx Let V(x) = kx^n ∴ A = E^(1/2 + 1/n) k^(-1/n) ∫ y^(1/n - 1) √(sin(π - y)) dy ⇒ A ∝ E^(1/2 + 1/n) ② A ball is dropped from a bridge 122.5 meters above a river. After the ball has been falling for 2 seconds, a second ball is thrown straight down after it. What must its initial velocity be so that both hit the water at the same time? Solution: time for first ball t₁ = √(2h/g) = √(2 * 122.5 / 9.8) = 5 sec time for second ball t₂ = 5 - 2 = 3 sec Now, s = ut + (1/2)gt² ⇒ 122.5 = u * 3 + (1/2) * 9.8 * 9 Problem 1 A body sliding on a smooth inclined plane requires 4 sec. to reach bottom after starting from rest at the top. How much time does it take to cover one-fourth the distance starting from rest at the top? Solution: s = ut + (1/2)at² ⇒ t = √(2s/a) t₁ ∝ √s t₂ ∝ √(s/4) = (1/2)√s t₂/t₁ = (1/2)√s / √s = 1/2 t₂ = t₁/2 = 4/2 = 2 sec Problem 2 A particle of mass m is placed on an inclined plane making an adjustable angle θ with horizontal as shown in the figure. The coefficient of friction between the particle and the inclined plane is μ. If the inclined plane is moving horizontally with a uniform acceleration a (see figure), the value of θ for which the particle will remain at rest on the plane is? Solution: Apply equilibrium of forces on y-axis: N cos θ - mg = 0 N = mg / cos θ Apply equilibrium of forces on x-axis: N sin θ - μN cos θ = ma (mg / cos θ) sin θ - μ (mg / cos θ) cos θ = ma tan θ - μ = a/g tan θ = (a + μg) / (g - μa) Problem: A block of mass 4 kg is attached to a spring of spring constant 100 N/m. One end of the spring is attached to the roof. Initially, the block is at rest. When released, the maximum speed attained by the block is: Solution: EOM: mg - kx = ma = m(dv/dt) = m(dv/dx) * (dx/dt) = m(dv/dx) * v ⇒ (mg - kx) dx = m(dv/dx) * v * dx ⇒ ∫(mg - kx) dx = ∫m * v * dv ⇒ mgx - (1/2)kx² = (1/2)mv² for max. speed, v is max ⇒ (dv/dt) is max ⇒ (dv/dx) is max (dv/dx) = (1/m) * (mg - kx) = 0 ⇒ x = mg/k v² = (2/m) * (mgx - (1/2)kx²) = (2/m) * (mg * (mg/k) - (1/2)k * (mg/k)²) = mg²/k v = √(mg²/k) = √((4 * 10) * (10)² / 100) = √(4 * 10) = 2 m/s