# Conservation of Momentum for explosion

1. Nov 1, 2008

### possum30540

1. The problem statement, all variables and given/known data
A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder had been placed on one of the blocks. The explosion does not harm the blocks but it doubles their total kinetic energy. After the explosion, the blocks move along the x axis and the orginally moving block has a speed of:

2. Relevant equations
momentum = mass * velocity

kinetic energy = (1/2) m *(v)^2

conservation on momentum: the total initial momentum = the total final momentum

3. The attempt at a solution
since the mass of the two cars is the same, the mass = M
v1 = velocity of cart 1
v2 = velocity of cart 2

before the collision, the total kinetic energy is 12.5M J
before the collision, the total momentum is 5M kg*m/s

after the collision, the total kinetic energy is 25M J

initial momentum = final momentum
5M = Mv1 + Mv2

This is where I am stuck. I don't know what the next step is and if someone would help me along it would be very much appreciated.

2. Nov 1, 2008

### HallsofIvy

Staff Emeritus
You say that the total kinetic energy after the collision is 25M J. Write the equation fror that!

3. Nov 1, 2008

### possum30540

The problem says the initial kinetic energy is doubled. So 2 * 12.5 = 25 J

4. Nov 2, 2008

### HallsofIvy

Staff Emeritus
Yes, you said that before. Now USE it! Write the equation for total kinetic energy after the collision-explosion and set it equal to 25.

5. Nov 2, 2008

### possum30540

25 J = (1/2)Mv1^2 + (1/2)Mv2^2