Momentum and explosion?

Question

Three smooth stones A, B, C are initially at rest, in contact with each other on the smooth surface of a frozen lake. The masses of stones are A = 800g B = 600g and C = 250g and the coefficient of sliding friction between the stones and the ice is 0.0250. 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 2.40 ms^-1. Stone B flies off due east with an initial speed of 3.60 ms^-1.

Q1 Calculate the speed and direction of motion of stone C
immediately after the explosion.
I got v = 11.56 ms^-1 and direction = 41.67 degrees.
I reckon that is south west? Any ideas.

Q2 If 10 per cent of the total energy of the explosion was converted to kinetic energy of the stones, what was the total energy released in the explosion
Q3 Show all the forces acting on stone A when sliding on
the ice after the explosion. How far does it travel before it
comes to rest. g = 9.81ms^-2

Answer 1

You're right with your answers to the first part. The angle is + 41.7 degrees which is in the bottom left quadrant.

The stones energy can be calculated from KE = 0.5*m*v^2 for each stone and simply added to give 22.9 J. This is 10% of the explosion's energy, which meant it released a total energy of 229 J.

The only force acting on A is that due to friction f=uN

Hence ma = uN = umg

a = ug = (0.025)*(9.8) = 0.245

This is acting to slow the stone down from a starting speed of u = 2.4 m/s

v^2 = u^2 + 2as
0 = 2.4^2 -2*0.245*s
s = 11.76 m