Conservation of linear momentum stationary block

[skaring]

I have been trying and trying to get this problem. I am using the fact that delta K = Fd, I set 0= to Fd of one part + Fd to the other part, and it is not the correct answer.

In Figure 9-57, a stationary block explodes into two pieces L and R that slide across a frictionless floor and then into regions with friction, where they stop. Piece L, with a mass of 2.4 kg, encounters a coefficient of kinetic friction µL = 0.40 and slides to a stop in distance dL = 0.15 m. Piece R encounters a coefficient of kinetic friction µR = 0.50 and slides to a stop in distance dR = 0.34 m. What was the mass of the original block?

 

[Doc Al]

I have been trying and trying to get this problem. I am using the fact that delta K = Fd, I set 0= to Fd of one part + Fd to the other part, and it is not the correct answer.

It sounds like you are trying to apply conservation of energy. But energy is not conserved--the block exploded!

But you can use \Delta KE = Fd for each piece to find the initial speeds of the two pieces just after the explosion. Do that and then use conservation of momentum to analyze the explosion itself.

 

[wais]

It seems like you have the right idea by using the conservation of linear momentum and the work-energy theorem, but there may be some mistakes in your calculations. Let's take a closer look at the problem.

First, let's define some variables:
m = mass of the original block
v = velocity of the original block before explosion
vL = velocity of piece L after explosion
vR = velocity of piece R after explosion

From the conservation of linear momentum, we know that the initial momentum of the block must equal the sum of the momenta of the two pieces after the explosion. This can be expressed as:

mv = m(vL + vR)

Next, we can use the work-energy theorem to relate the change in kinetic energy to the work done by friction. In this case, the work done by friction is equal to the force of friction multiplied by the distance traveled. This can be expressed as:

ΔK = W = Fd

Now, we can use the given information to set up two equations:

For piece L:
ΔKL = FdL = (µLmg)(dL) = 0.4mg(dL) = 0.4m(vL)^2

For piece R:
ΔKR = FdR = (µRmg)(dR) = 0.5mg(dR) = 0.5m(vR)^2

We can then substitute these equations into our initial equation from the conservation of linear momentum:

mv = m(vL + vR)

And rearrange for the velocity of the original block:
v = vL + vR

Now, we can substitute this into our equations for the change in kinetic energy:

ΔKL = 0.4m(v-vR)^2
ΔKR = 0.5m(v-vL)^2

We can then solve for v by setting these two equations equal to each other and solving for v:

0.4m(v-vR)^2 = 0.5m(v-vL)^2
0.4(v-vR)^2 = 0.5(v-vL)^2
0.4v^2 - 0.8vvR + 0.4vR^2 = 0.5v^2 - 0.5vvL - 0.5vL^2
0.1v^2 - 0.3vvR