Question

In the figure below, block 1 of mass m1 slides along an x axis on a frictionless floor with a speed of v1i = 3.20 m/s. Then it undergoes a one-dimensional elastic collision with stationary block 2 of mass m2 = 0.250m1. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 = 0.250m2. (a) Express the speed of block 3 just after the collision in terms of the initial speed of block 1, v1i. Note whether it is greater than, less than, or equal to the initial speed of block 1. v3 = ✕ v1i (b) Express the kinetic energy of block 3 just after the collision in terms of the initial kinetic energy of block 1, K1i. Note whether it is greater than, less than, or equal to the initial kinetic energy of block 1. K3 = ✕ K1i (c) Repeat for the magnitude of momentum of block 3 just after the collision, noting whether it is greater than, less than, or equal to the initial magnitude of momentum of block 1. p3 = ✕ p1i

Answer 1

Date (9) Collision between mi and me Apply Conservation of momentum mivi & mavi a mivy + mary mirii + o - mivet marry For elastic collision Val - Viq = vii - using 6 and o mivi a milv2f-vil & mary amivii = (mit m2) Vt Vet - amivii amri tigre - mitme mit 0.25m, For Collision blwma and my Vzt = 1.6rii / - Apply Consuration of momentum ma Vai + ma voi = my Vad + m₂ Vzt Here vai = Voeg 106 vi and Vizo ( - m2 (1060) = m₂ Vzf + maraf 3 for elastre collision Vay Vaya Vai Vaga Vaga loovid using ② and a 106 mavii - m2 (V3 - 1o6v, i) + m₂ Vof 2x 106 marii = (ma+ m₂) of - Vy= 3.2 mr vid 3.200.25m, vii met my 02rm, t 0.25% 0.25m, - vay = 2.56 Vid greater then 6 Kz at mg (Vat) 2 = + (0.25% 0.25m,) (2.560 4/² | K3= 3.28 ki greater than Sweethavat = 0.25% 0.25 m, (3.5% ) Selfie Dal6 less than O