Question

In the figure below, block 1 of mass m1 slides along an x axis on a frictionless floor with a speed of 1.80 m/s. Then it undergoes a one-dimensional elastic collision with stationary block 2 of mass m2 = 2.50m1. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 = 2.50m2. (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

(a) Velocity of m, after first collision is - 211 ni 2fm+ m2 2m2 V 3.50m = 0.57 Velocity of m, after second collision is, 2m, Vasm - 127 ,+m 2m2 (0.577) 3.50m2 = 0.3250, Vag=0.325(1.81) = 0.58825 m/s =0.60 m/s V37 <VI The final kinetic energy of block 3 is KE; = m;vis? = 3 (5.0m. )(0.325v) = 0.5281(3mi) = 0.5281KE KE; <KE (d) The final momentum is P = mys = (5.0m (0.325v;) = 1.63m Vi P, > P