1. The problem statement, all variables and given/known data Three smooth stones, A, B and C are initially at rest, in contact with each other on the smooth surface of a frozen lake (Figure 1). The masses of the stones are, to 3 significant figures, mA = 8.00 × 10^2 g, mB = 6.00 × 10^2 g and mC =2.50 × 10^2 g and the coefficient of sliding friction between the stones and the ice is μslide = 2.50 × 10^−2. An explosion between the stones causes them to fly apart across the surface of the ice. Stone A flies off due north with an initial speed of 2.40 m s^−1. Stone B flies off due east with an initial speed of 3.60 m s^−1. (a) Calculate the speed and direction of motion of stone C immediately after the explosion. You can neglect the effects of any exhaust material from the explosion. (Remember that this is a two-dimensional problem and that you should define a suitable coordinate system for your calculation using diagrams where necessary.) (b) If 10% of the total energy of the explosion was converted to the kinetic energy of the stones, what was the total energy released in the explosion? c) Draw a diagram showing all the forces acting on stone A when it is sliding across the ice after the explosion. Calculate how far it travels before it comes to rest. You can assume that the stone does not roll at any time. 2. Relevant equations a) momentum = mass x velocity Pythagoras c^2 = a^2 + b^2 Initial momentum = final momentum b) Kinetic = 1/2 x mass x velocity^2 c) F=ma where a = μ x g V = u + at s = ut + 1/2a(t^2) 3. The attempt at a solution a) Momentum A = 0.8 x 2.4 = 1.92 Momentum B = 0.6 x 3.6 = 2.16 Momentum C = 0.25 x V Using a vector triangle shown in attached picture under a) initial momentum = final momentum 0 = momentum A + momentum B - momentum C momentum C = momentum A+ momentum B From vector triangle (Mc x Vc)^2 = 1.92^2 + 2.16^2 Vc = (1.92^2)+(2.16^2) / (0.25^2) (m = 0.25) Vc = 11.6ms-1 Angle Rho = tan^-1 (1.92/2.16) = 41.6 degrees So the correct direction is is 138.4 degrees below the x axis b) Kinetic energy of each disc using 1/2 m v^2 0.5 x 0.8 x 2.4^2 = 2.3 0.5 x 0.6 x 3.6^2 = 3.9 0.5 x 0.25 x 11.6^2 = 16.7 Total energy given in explosion is sum of totals x 10 = 228.6 Joules c) Using diagram under b) Va = 2.4 acceleration = 9.81 x μ V = u + at 2.4= 0 + 0.25t t= 2.4/0.25 = 9.6 s s = ut + 0.5 a t^2 s = 0 + 0.5 x 0.25 x 9.6 s = 11.5m Please can you help. I struggle with this question.