Question

avg Plm for each time step. Use the approximation that v A paddle ball toy consists of a flat wooden paddle and a small rubber ball that are attached to each other by an elastic band (figure). You have a paddle ball toy for which the mass of the ball is 0.014 kg, the stiffness of the elastic band is 0.925 N/m, and the relaxed length of the elastic band is 0.350 m. You are holding the paddle so the ball hangs suspended under it, when your cat comes along and bats the ball around, setting it in motion. At a particular instant the momentum of the ball is <-0.02, -0.01, -0.02kg.m/s, and the moving ball is at location <-0.2, -0.61, 0> m relative to an origin located at the point where the elastic band is attached to the paddle

Answer 1

The initial momentum is, Piritid = (-0.02i – 0.01j -0.02k) kg.m/s The initial position is, l'initial = (-0.2i - 0.61j + Ok) m The magnitude of position is, Pininiai = (-0.2) +(-0.61) = 0.64 m Using newton's second law, (0.925) (0.64 -0.35)(-0,21 -0.61j - Ok) Fever =(-0.014) (9.81);- = -0.13734j-(0.41914)(-0.21-0.61j+0k) = -0.137347 +0.0838i +0.25567j = (0.083&i +0.118j) N 0.64 (a) The conservation of momentum is, Promar = Pinitia: + FreAt =(-0.02 -0.01-0.02k)+(0.0838i +0.118j)(0.1) =(-0.021 -0.01% -0.02k)+(0.00838 +0.00118j) = -0.011621 -0.00882j -0.02k The aveage velocity is, m -0.01 1621 -0.00882j -0.02k 0.014 =-0.831-0.63)-1.429k ("forai= lininiai + Vang (At) =(-0.2 -0.61j + Ok)+(-0.83i-0.63; -1.429k)(0.1) =-0.283 -0.673; -0.143k (b) The conservation of momentum is, Pre = P + FA =(-0.021 -0.01-0.02k)+(0.08381 +0.118j)(0.1) =(-0.02 -0.01j-0.02k)+(0.00838 +0.00118j) =-0.01162 -0.008823 -0.02k

The aveage velocity is m -0.01 1621 -0.00882j -0.02k 0.014 = -0.83i-0.63; -1.429k ("final ) = limitiai + Veng (AT) =(-0.21 -0.61j+Ok)+(-0.83i -0.63; -1.429k)(0.05) =-0.242i - 0.642j -0.0715k