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

2024
Views:

439

Pages:

5
Author:

Jabin F.

#Equation of Path (Relation b/w co-ordinates) From eq (1) & (2) [y = ux\sinθ - \frac{g}{2}x²] [y = x\tanθ - \frac{g}{2u²cos²θ}x²] For time of flight: Put t = T & y = 0 in eq (2) [0 = uT\sinθ - \frac{g}{2}T²] [T = \frac{2usinθ}{g}] Therefore, range: R = Put t = T & x = R in eq (1) [R = uT\cosθ = u \cdot \frac{2usinθ}{g} \cdot \cosθ] [R = \frac{2u²sinθcosθ}{g} = \frac{u²sin2θ}{g}] For max. height: H Put t = T/2 & y = H in eq (2) [H = u(\frac{T}{2})\sinθ - \frac{g}{2}(\frac{T}{2})² = \frac{u}{2} \cdot \frac{2usinθ}{g} \cdot \sinθ - \frac{g}{2} \cdot \frac{4u²sin²θ}{4g²}] [H = \frac{u²sin²θ}{2g}] Note: When the particle is thrown with a large speed, then it will go to a higher height. Comparison with the earth: If it is near to the earth, then the path will be an ellipse, not a parabola. P-57 A ball is dropped vertically from a height H onto a plane surface and permitted to bounce vertically repeatedly along a vertical line. After every bounce, its kinetic energy becomes a quarter of its kinetic energy before the bounce. The ball will come to rest after time? Solution: For first bounce, ball is drop from height H in gravity, therefore the velocity of ball just before the bounce is [v = \sqrt{2gH}] As according to question the kinetic energy decreases, so, if v₁ is velocity before bounce and v₂ is velocity after 1st bounce then, (KE) before bounce = (KE) after bounce [\frac{1}{2}mv₁² = \frac{1}{4}(\frac{1}{2}mv₁²)] [v₂² = \frac{9gH}{4} \Rightarrow v₂ = \frac{\sqrt{9gH}}{2} = \frac{3\sqrt{gH}}{2}] Similarly on 2nd bounce v₃ = (3/4)v₂ = (3/4)(3√(gH)/2) = (9/8)√(gH) Let t be the time taken for 1st bounce [t₁ = \frac{v₁}{g} = \frac{\sqrt{2gH}}{g} = \sqrt{\frac{2H}{g}}] Similarly t₂ be the time b/w 1st & 2nd collision: [t₂ = \frac{v₂}{g} = \frac{3\sqrt{gH}}{2g} = \frac{3}{4}\sqrt{\frac{2H}{g}} = \frac{3}{4}t] Similarly t₃ be the time b/w 2nd & 3rd collision: [t₃ = \frac{v₃}{g} = \frac{9}{8}\sqrt{\frac{gH}{g}} = \frac{9}{8}t] Then total time T will be: [T = t + t₁ + t₂ + t₃ + ... = t + t + \frac{t}{4} + \frac{t}{8} + ...] This forms an infinite GP with a = t and r = 1/4 ∴ Sum of this GP will be [S = \frac{a}{1-r} = \frac{t}{1 - (1/4)} = \frac{4t}{3}] [T = \frac{4}{3}\sqrt{\frac{2H}{g}}] [T = t[1 + \frac{1}{4} + \frac{1}{16} + ...] = t[1 + \frac{1}{3}] = \frac{4t}{3} = \frac{4}{3}\sqrt{\frac{2H}{g}}]