Dynamics Chapter 5 Impact and Collision PDF

# Chapter (5) Impact & Collision ## Work - `dw = (F cos θ) ds = F . ds` - `W = ∫ F . ds` - `W = F . PP` (Joule or erg) - If F is constant vector ## Energy - `F = m f` => `m dv/dt = F` - `∫ m dv/dt = ∫ F dx` (If it starts from rest at P₀) - `∫ m v dv = ∫ F dx` - `½ m v²...

# Chapter (5) Impact & Collision ## Work - `dw = (F cos θ) ds = F . ds` - `W = ∫ F . ds` - `W = F . PP` (Joule or erg) - If F is constant vector ## Energy - `F = m f` => `m dv/dt = F` - `∫ m dv/dt = ∫ F dx` (If it starts from rest at P₀) - `∫ m v dv = ∫ F dx` - `½ m v<sup>2</sup> = ∫ F dx` = Work - K.E = ½ m v² = ∫ F dx - `V = ∫ F dt` = `-∫ F dx` - P.E = `-∫ mg dx = mg h` ( If the particle at height h above the earth's surface) ## Principle of work as equation (4) - `∫ m v dv = ∫ F dx` - `½ m v<sub>2</sub><sup>2</sup> - ½ m v<sub>1</sub><sup>2</sup> = ∫ F dx` - Change in K.E = Work done ## Conservation of Energy - `T<sub>2</sub> - T<sub>1</sub> = ∫ F dx + ∫ F dx` - `T<sub>2</sub> - T<sub>1</sub> = V<sub>2</sub> - V<sub>1</sub>` - `T<sub>2</sub> + V<sub>2</sub> = T<sub>1</sub> + V<sub>1</sub>` => `T + V = Constant` ## Impulse - `I = F (t<sub>2</sub> - t<sub>1</sub>)` (If F is constant vector) - If F is variable - `I = ∫ F(t) dt = ∫ m dv/dt dt = ∫ m dv` - `I = m v<sub>2</sub> - m v<sub>1</sub>` (Change in momentum) - Impulse <--------- ---------> ## Conservation of Momentum - By Newton's 3rd law - `R<sub>1</sub> = - R<sub>2</sub>` - `m<sub>1</sub> u<sub>1</sub> + m<sub>2</sub> u<sub>2</sub> = m<sub>1</sub> v<sub>1</sub> + m<sub>2</sub> v<sub>2</sub> ` ## Newton's experimental Law: - Along the direction of line of impact - `v<sub>2</sub> = v<sub>1</sub> - e (v<sub>2</sub> - v<sub>1</sub>)` - V₂ - v₁ = e (v₂ - v₁) (Before Collision) - G_e = e (Coefficient of restitution) - G_e = 1 (Perfectly elastic body) - 0 < G_e < 1 (for inelastic bodies) - Types of impact: - Direct - Oblique ## (I) Direct impact of two spheres - Before impact - After impact - `m<sub>1</sub> u<sub>1</sub> + m<sub>2</sub> u<sub>2</sub> = m<sub>1</sub> v<sub>1</sub> + m<sub>2</sub> v<sub>2</sub>` - `u<sub>1</sub> - u<sub>2</sub> = -e (v<sub>1</sub> - v<sub>2</sub>)` Solve 1, 2, we get: - `v<sub>1</sub> = ` - ` v<sub>2</sub> = ` ## Example (I) - Three perfectly elastic spheres A, B, C of masses 7m, 7m, m respectively are at rest with their centers along a straight line and C lies between A and B. - The sphere C is projected towards A with velocity u. - Prove that three collisions only occur between the three spheres and the ratio between the final velocities is 21:12:1 1. **1st Collision bet. C,A** - `m<sub>1</sub> u<sub>1</sub> + 7m<sub>2</sub> u<sub>2</sub> = 7m<sub>1</sub> v<sub>1</sub> + 7m<sub>2</sub> v<sub>2</sub>` - `v<sub>2</sub> - u = -e (0 - u)` ; e = 1 - `-v<sub>1</sub> + v<sub>2</sub> = u ` - `v<sub>2</sub>` = `4u/7` - `v<sub>1</sub>` = `-3u/7` - `-ve` sign means that C will move in the opposite direction towards B and Collision will occur. 2. **2nd collision bet. C,B** - `m<sub>1</sub> u<sub>1</sub> + 7m<sub>2</sub> u<sub>2</sub> = 3m<sub>1</sub> v<sub>1</sub> + 3m<sub>2</sub> v<sub>2</sub>` - `v<sub>2</sub> - 4u/7 = - e (0 - 4u/7)` ; e = 1 - `-v<sub>1</sub> + v<sub>2</sub> = 4u/7` - `v<sub>2</sub>` = `16u/21` - `v<sub>1</sub>` = `8u/21` - Thus C will reverse its direction and Collides with A again. 3. **Collision bet. C,A** - `v<sub>c</sub>` = `7u/16` - `v<sub>A</sub>` = `-9u/ 16` - `v<sub>2</sub> - v<sub>1</sub> = -e (u - v<sub>1</sub>)` ; e = 1 - `v<sub>2</sub> - v<sub>1</sub> = (u - v<sub>1</sub>)/2` - `-v<sub>1</sub> + v<sub>2</sub> = 37u/ 16` - `v<sub>1</sub>` = `-3u/64` - ` v<sub>2</sub>` = `21u/64` - `v<sub>A</sub>` = `-3u/64` - Thus the final velocities: - `v<sub>B</sub>` = `-8u/21` - `v<sub>c</sub>` = `64u/64` - `v<sub>A</sub>` = `21u/64` - Hence there are only **three collisions occur** - And then: `v<sub>A</sub>` : `v<sub>B</sub>` : `v<sub>c</sub>` = (21u/64) : (12u/64) : (u/64) = 21:12:1 ## (II) Oblique impact - Between two spheres: - `v<sub>1</sub> = ` - `v<sub>2</sub> = ` - Since there is no impulsive force in the direction (attr) perpendicular to the line of impact (Centers), the vertical Component's of velocities unchanged. - So the horizontal `v<sub>1</sub>` , `v<sub>2</sub>`, `v<sub>1</sub>` , `v<sub>2</sub>` can be evaluated as before. - `m<sub>1</sub> u<sub>1</sub> + m<sub>2</sub> u<sub>2</sub> = m<sub>1</sub> v<sub>1</sub> (G<sub>1</sub>cosθ<sub>1</sub>) + m<sub>2</sub> v<sub>2</sub> (G<sub>2</sub>cosθ<sub>2</sub>)` - `v<sub>2</sub> - v<sub>1</sub> = ` - ` ½ m<sub>1</sub> u<sub>1</sub><sup>2</sup> + ½ m<sub>2</sub> u<sub>2</sub><sup>2</sup> = ½ m<sub>1</sub> v<sub>1</sub><sup>2</sup>(G<sub>1</sub>cosθ<sub>1</sub>)<sup>2</sup> + ½ m<sub>2</sub> v<sub>2</sub><sup>2</sup> (G<sub>2</sub>cosθ<sub>2</sub>)<sup>2</sup>` - When `u<sub>1</sub> = u<sub>1</sub>, u<sub>2</sub> = u<sub>2</sub>, e = ½, ` α₁ = α₂, tan α₁ = `-l/2`. ## Example - A sphere of mass 5m and moving with velocity 13ft/s impinges another sphere of mass m that moves with velocity 5ft/s, their directions of motion being inclined to the line of centers at angles Sin^-1(5/13) and Sin^-1(3/5) respectively. If the coefficient of restitution is ½, - Find the magnitude and direction of their velocities after impact - The impulsive reaction between the two spheres - The loss in Kinetic energy. ## (II₂) Between sphere and plane - The velocity of the sphere // tl to the plane is invariant. - Here we don't know the mass of the plane, hence applying Newton's exp. law `v<sub>1</sub> - D = -e (-v<sub>1</sub>G<sub>1</sub>) => (I = e v<sub>1</sub>G<sub>1</sub>)` i.e `v<sub>1</sub>`C = `-e` v₁G₁ - `v<sub>1</sub><sup>2</sup>` = `v<sub>1</sub><sup>2</sup>sin<sup>2</sup>θ + e<sup>2</sup>u<sup>2</sup>cos<sup>2</sup>θ` = `u<sup>2</sup>(sin<sup>2</sup>θ + e<sup>2</sup>cos<sup>2</sup>θ)` - `tan<sup>2</sup>α` = `e<sup>2</sup>tan<sup>2</sup>θ` - V₁ = `e tan θ` ***