Question

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?

Answer 1

Let the initial velocity of L and R be u_L and u_R respectively, then by conservation of momentum,

m_Lu_L = m_Ru_R

For L:

acceleration, a= -_Lg; final velocity, v_L=0; s_L=0.41 m

using, v_L² - u_L² = 2as

u_L= 2.0452 m/s

For R:

acceleration, a= -_Rg; final velocity, v_R=0; s_R=0.44 m

using, v_R² - u_R² = 2as

u_R= 1.687 m/s

putting the values of u_L and u_R and m_Lin the first equation,

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

m_R= 2.6671 kg

mass of the block, m= 2.2+2.66

m= 4.86 kg