COM O21

Problem Statement: A plank of mass 1.5 kg is placed on a lubricated floor with viscous drag $F = -kv$. A block of mass 0.5 kg lands on it with velocity 10 m/s. Find the total distance the plank slides.

Analysis:

Let $m = 0.5$ kg be the mass of the block and $M = 1.5$ kg be the mass of the plank. The drag coefficient is $k = 2.0$ kg/s.

We apply the Impulse-Momentum theorem to both bodies over the entire duration of motion (from $t=0$ until both stop).

1. Impulse on the Plank:

The plank starts at rest and ends at rest. The forces acting on it in the horizontal direction are the forward friction $f_k$ from the block and the backward viscous drag $kv$.

$$ \Delta P_{plank} = \int (f_k – kv) \, dt = M(v_f – v_i) = 0 $$ $$ \int f_k \, dt = k \int v \, dt $$

Since $\int v \, dt$ is the total displacement $S$:

$$ \int f_k \, dt = kS \quad \text{— (i)} $$

2. Impulse on the Block:

The block starts with $v_0 = 10$ m/s and stops. The only horizontal force is friction $f_k$ acting backwards.

$$ \Delta P_{block} = \int (-f_k) \, dt = m(0 – v_0) $$ $$ -\int f_k \, dt = -mv_0 $$ $$ \int f_k \, dt = mv_0 \quad \text{— (ii)} $$

3. Calculation:

Equating (i) and (ii):

$$ kS = mv_0 $$ $$ S = \frac{mv_0}{k} $$

Substituting the values:

$$ S = \frac{0.5 \times 10}{2.0} = \frac{5}{2} = 2.5 \text{ m} $$