Conservation of Momentum Involving Friction

[Goatsenator]

Homework Statement

In the figure here, 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 1.6 kg, encounters a coefficient of kinetic friction μL = 0.43 and slides to a stop in distance dL = 0.42 m. Piece R encounters a coefficient of kinetic friction μR = 0.39 and slides to a stop in distance dR = 0.48 m. What was the mass of the block?

Homework Equations

Pi = PLf + PRf

The Attempt at a Solution

I started off assuming EmecR1 - fkdr = EmecR2 [no U involved]

then just after the explosion 1/2(mR)(vRi)^2 - (mR)(0.39)(g)(dr) = 0 [it comes to a stop]

I pull out mR and divide each side by mR to get 1/2(vRi)^2 - (.39)(g)(dr) = 0

then solve for vRi and I'm getting 1.9155 m/s


Next I do the same procedure for the Left block

1/2(1.6)(vLi)^2 - (1.6)(.43)(9.8)(.42) = 0

then vLi = -1.8814 [b/c block L is traveling in the negative x direction]


plugging into the conservation of momentum

Pi = 0 = (1.6)(-1.8814) + (mR)(1.9155)

then I get mR = 1.572 Kg but that's not right. Can anyone help me with this?

 

[cepheid]

Hi Goatsenator,

I don't see anything wrong with your physics or with your arithmetic. It could just be something really stupid: the problem asks for you to solve for the mass of "the block." Presumably this means the original block (before it exploded). So you'd have to add the answer you got to 1.6 kg to get the total block mass.

 

[Goatsenator]

Oh man! I keep making these stupid mistakes! -_- I guess they could have been a little clearer. Oh well. Yeah that was right thank you for the help!

 

[cepheid]

Oh man! I keep making these stupid mistakes! -_- I guess they could have been a little clearer. Oh well. Yeah that was right thank you for the help!

You're welcome!

 

[rwisz]

Your approach to solving this problem is correct, however there seems to be an error in your calculation for the mass of the block. The correct solution is as follows:

First, we can use the conservation of momentum equation, Pi = PLf + PRf, to find the initial momentum of the block before it explodes. Since the block is initially at rest, the initial momentum is equal to zero.

Next, we can use the equations for the final momentum of the two pieces L and R, which are given by PLf = mLvLf and PRf = mRvRf. Substituting these equations into the conservation of momentum equation, we get:

0 = mLvLf + mRvRf

Next, we can use the equations for the final velocity of each piece, which are given by vLf = vLi + aLt and vRf = vRi + aRt. Here, aL and aR represent the acceleration of each piece due to friction, and t represents the time it takes for each piece to come to a stop. Substituting these equations into the previous equation, we get:

0 = mL(vLi + aLt) + mR(vRi + aRt)

Since both pieces come to a stop, their final velocities are equal to zero, so we can simplify the equation to:

0 = mL(vLi + aLt) + mR(vRi + aRt) = -mLaLdL - mRaRdR

We can then solve for the mass of the block using the given values for aL, aR, dL, and dR:

mL = 1.6 kg
aL = μLg = 0.43(9.8 m/s^2) = 4.214 m/s^2
dL = 0.42 m
mR = m (unknown)
aR = μRg = 0.39(9.8 m/s^2) = 3.822 m/s^2
dR = 0.48 m

Plugging in these values and solving for m, we get:

0 = -1.6(4.214)(0.42) - m(3.822)(0.48)
m = 1.6(4.214)(0.42) / (3.822)(0.48)
m