COM BYU 13

System Analysis (Cart Frame):

Consider the motion of block A relative to cart B. The block starts at the left end, travels distance $l$ to the right, hits the obstruction elastically, and returns distance $l$ to the left end (rear) where it stops.

Step 1: Relative Acceleration

Friction $f = \mu mg$ acts on both blocks (action-reaction).
Acceleration of A (ground): $a_A = -\mu g$.
Acceleration of B (ground): $a_B = \frac{\mu mg}{M}$.
Relative acceleration (retarding):

$$ a_{rel} = |a_A – a_B| = \mu g + \frac{\mu mg}{M} = \mu g \left( \frac{M+m}{M} \right) $$

Step 2: Time Calculation

The total relative distance covered is $S_{rel} = 2l$. The block starts with relative velocity $u_{rel}$ and ends with 0.

Using kinematics $S = \frac{1}{2} a_{rel} t^2$ (considering the motion in reverse from stop to start):

$$ t = \sqrt{\frac{2 S_{rel}}{a_{rel}}} = \sqrt{\frac{2(2l)}{\mu g \frac{M+m}{M}}} = 2 \sqrt{\frac{Ml}{\mu g(M+m)}} $$

Substituting values ($M=8, m=2, l=1, \mu=0.5$):

$$ t = 2 \sqrt{\frac{8(1)}{0.5(10)(10)}} = 2 \sqrt{\frac{8}{50}} = 2(0.4) = 0.8 \text{ s} $$

Step 3: Loss of Mechanical Energy

The loss of mechanical energy corresponds to the work done by friction against the relative sliding.

$$ \Delta E = f \cdot S_{rel} = (\mu mg) \cdot (2l) $$ $$ \Delta E = 2 \mu m g l $$