# Speed and coefficeint of friction part 2

1. Apr 8, 2007

### imy786

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. The masses of the stones are A = 400 g, B = 300 g and C = 120 g and the coefficient of sliding friction between the stones and the ice is μslide = 0.0150. 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 4.50 m s−1. Stone B flies off due east with an initial speed of 6.60 m s−1.

If 20% 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?

2. Relevant equations

KE= 1/2 * M * V^2

PE= mgh

3. The attempt at a solution

total energy= ke+ pe

work done= force*distance

force= uR

μslide = 0.0150

i dont know R. And i dont know the distance travelend to calculate work done.

Need help please.

2. Apr 8, 2007

### Meir Achuz

1. Use cons of momentum to find the momentum and then the KE of stone C.
2. The KE of A + B + C equals 1/5 of the energy in the explosion.
You don't use PE, work done, or frliction.

3. Apr 8, 2007

### imy786

why have they given friction in the quesiton at all??
i didnt use friction in the first part either.

4. Apr 8, 2007

### robb_

I dont see a need for it either. (here)

5. Apr 10, 2007

### Meir Achuz

Maybe it is a smokescreen to make you realize that friction plays no part in collisions.

6. May 10, 2007

### imy786

stone C= m1u1 + m2u2 = m1v1 +m2v2

m1= 0.120 kg
u1= 0

m2= would it stone B + stonce c mass
u2=? i got no idea what value to put here, as i have the speed of stone b and c.

could i not just calculate the total energy by just calculating KE , or do i have to also calculate conversation of momentum?

7. May 11, 2007

### Meir Achuz

Your first equation is not relevant here.
Start with momentum conservation
$$0=m_A{\vec v_A}+m_B{\vec v_B}+m_C{\vec v_C}.$$
to find v_c. Then use energy.

8. May 11, 2007

### imy786

https://www.physicsforums.com/showthread.php?t=161727

speed of c is aprox 36meters/sec in the south west direction.

KE after collison= 0.5mava^2 + 0.5Mbvb^2 + 0.5McVc^2
= 0.5 ( 0.4 * 4.5^2 + 0.3*6.6^2 + 0.120* 36^2)
= 0.5 (8.1+ 13.068+ 155.52)
=0.5* 176.688
=88.3 J

9. May 15, 2007

### imy786

is this correct to for the calculation to work out KE

10. May 15, 2007

### hage567

Are you sure that's the value you found for the speed of stone C? Looking at the other thread, I think you determined you had made a mistake with that. Double check it.

11. May 16, 2007

### imy786

is the method correct though to work it calculating KE

12. May 16, 2007

### hage567

It's right for finding the total kinetic energy after the explosion.

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