Question

# In the figure here, a stationary block explodes into two pieces L and R that slide...

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 2.2 kg, encounters a coefficient of kinetic friction μL = 0.52 and slides to a stop in distance dL = 0.41 m. Piece R encounters a coefficient of kinetic friction μR = 0.33 and slides to a stop in distance dR = 0.44 m. What was the mass of the block?

Let the initial velocity of L and R be uL and uR respectively, then by conservation of momentum,

mLuL = mRuR

For L:

acceleration, a= - Lg; final velocity, vL=0; sL=0.41 m

using, vL2 - uL2 = 2as

uL= 2.0452 m/s

For R:

acceleration, a= - Rg; final velocity, vR=0; sR=0.44 m

using, vR2 - uR2 = 2as

uR= 1.687 m/s

putting the values of uL and uR and mLin the first equation,

2.2*2.0452 - mR*1.687 = 0 {negative sign indicates that the velocities are in the opposite direction}

mR= 2.6671 kg

mass of the block, m= 2.2+2.66

m= 4.86 kg

#### Earn Coins

Coins can be redeemed for fabulous gifts.