Answer
(a) The speed of the brick after the collision is $0.0476~v_0$
(b) 95.2% of the mechanical energy was lost in the collision.
Work Step by Step
Let $m_c$ be the mass of the clay.
Let $m_b$ be the mass of the brick.
(a) We can use conservation of momentum to set up an equation for this perfectly inelastic collision.
$p_f=p_0$
$(m_c+m_b)~v_f = m_c~v_0$
$v_f = \frac{m_c~v_0}{m_c+m_b}$
$v_f = \frac{(0.050~kg)~v_0}{0.050~kg+1.0~kg}$
$v_f = 0.0476~v_0$
The speed of the brick after the collision is $0.0476~v_0$
(b) We can find the initial kinetic energy.
$KE_0 = \frac{1}{2}m_cv_0^2$
$KE_0 = \frac{1}{2}(0.050~kg)v_0^2$
$KE_0 = 0.025~v_0^2$
We can find the final kinetic energy.
$KE_f = \frac{1}{2}(m_c+m_b)(0.0476~v_0)^2$
$KE_f = \frac{1}{2}(0.050~kg+1.0~kg)(0.0476~v_0)^2$
$KE_f = 0.0012~v_0^2$
We can find the percentage of kinetic energy that remains after the collision.
$\frac{KE_f}{KE_0}\times 100\% = \frac{0.0012~v_0^2}{0.025~v_0^2}\times 100\% = 4.8\%$
We can find the percentage of kinetic energy that was lost in the collision.
$100\% - 4.8\% = 95.2\%$
95.2% of the mechanical energy was lost in the collision.
