Answer
The mass of the block was $~~3.4~kg$
Work Step by Step
Let $m_L$ be the mass of piece L
We can find the magnitude of deceleration of piece L:
$F = m_L~a$
$m_L~g~\mu_L = m_L~a$
$a = g~\mu_L$
$a = (9.8~m/s^2)(0.40)$
$a = 3.92~m/s^2$
We can find the speed of piece L as it enters the region with friction:
$v_f^2 = v_0^2+2ax$
$0^2 = v_0^2+2ax$
$v_0^2= -2ax$
$v_0= \sqrt{-2ax}$
$v_0= \sqrt{-(2)(-3.92~m/s^2)(0.15~m)}$
$v_0 = 1.084~m/s$
Let $m_R$ be the mass of piece R
We can find the magnitude of deceleration of piece R:
$F = m_R~a$
$m_R~g~\mu_R = m_R~a$
$a = g~\mu_R$
$a = (9.8~m/s^2)(0.50)$
$a = 4.9~m/s^2$
We can find the speed of piece R as it enters the region with friction:
$v_f^2 = v_0^2+2ax$
$0^2 = v_0^2+2ax$
$v_0^2= -2ax$
$v_0= \sqrt{-2ax}$
$v_0= \sqrt{-(2)(-4.9~m/s^2)(0.25~m)}$
$v_0 = 1.565~m/s$
By conservation of momentum, the magnitude of each piece's momentum after the explosion must be equal. We can find $m_R$:
$(m_R)(1.565~m/s) = (m_L)(1.084~m/s)$
$m_R = \frac{(2.0~kg)(1.084~m/s)}{1.565~m/s}$
$m_R = 1.4~kg$
We can find the mass of the block:
$m_R+m_L = 1.4~kg+2.0~kg = 3.4~kg$
The mass of the block was $~~3.4~kg$
