Three Body 2D Explosion with Friction Problem

[ford666]

1. The problem statement, all variables and given/known data

There are three smooth stones in contact with each other at rest on an ice field (coefficient of sliding friction=0.0250). Their masses are: A = 0.800kg, B = 0.600kg and C = 0.250kg. An explosion causes them to fly apart. 'A' moves Due North with a Velocity = 2.40 ms-1, 'B' moves Due East with a Velocity = 3.60 ms-1 .

Calculate speed & direction of stone C.
Show all forces acting on Stone A

2. Relevant equations
I can't find an equation which combines momentum and coefficient of friction.

Fr = coefficient of sliding Friction x N
P =mv, KE = 1/2 mv2
F = ma

3.
The attempt at a solutionI thought this was a simple momentum problem, apart from finding the direction of C, but the addition of the coefficient of friction has thrown me. I don't have a time scale so I can't use Force vectors as I can't get the accelerations. Please help, how do I approach this?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[ford666]

Calculate Velocity of Stone C

Known Information:

M a = 0.800 kg
M b = 0.600 kg
M c = 0.250 kg
V a = 2.40 ms-1
V b = 3.60 ms-1
V c = ?

KE = ½ MV2; P = MV

P a = M a V a P b = M b V b P c = M c V c


M c V c = √ (M a V a )2 (M b V b)2


V c = √ (M a V a )2 (M b V b)2 / M c



V c = √(0.800 kg x 2.40 ms-1)2 x (0.600 kg x 3.60 ms-1)2 / 0.250 kg

V c = √3.6864 x 4.6656 kg ms-1 / 0.250 kg


V c = √17.19927 / 0.250 kg


V c = 4.1472 / 0.250 = 16.5880 ms-1


Calculating Direction:

Tan θ˚ = P b
P a

Tan θ˚ = 3.6864
4.6656

Tan θ˚ = 0.81834

θ˚ = -38.295˚ with respect to x-axis.

Velocity & Direction of C = 16.5880 ms-1 @ -128.295˚ (SW)

Is this the correct method?

(b)
Calculating Energies

KEexp = KEa + KEb + KEc

KEexp = ½ M a V a2 + ½ M b V b2 + ½ M c V c2

KEexp = ½ x 0.800 kg x (2.40)2 + ½ 0.600 kg x (3.60)2 + ½ x 0.250 kg x (16.5880)2

KEexp = 2.304 + 3.888 + 33.162 J

KEexp = 39.3543 J

Etotal = KEexp x 10

Etotal = 393.543 J

Etotal = 394 J to 3 significant figures.


(c)

Forces acting on Stone A:

Force of Explosion – Kinetic Energy transferred to Stone producing Velocity 16.5880 ms-1 -128.295˚ (SW). Force of Friction of Stone on Ice working against the direction of motion. Weight of the Stone on the ice (M c g) in the direction of gravity. After initial velocity speed reduces to zero over time (t). Distance travelled by stone = | V | x (t).

How do I calculate the Distance without the (t) s.