Problem 2: Block on a Plank in Water
- Impulse and Momentum: The change in momentum is equal to the impulse of the external force ($\Delta p = \int F dt$).
- Conservation of Momentum (Internal): Momentum is conserved during the inelastic collision between the block and the plank.
- Variable Force Integration: Integration of velocity-dependent force over distance.
Step 1: Analyze the motion and forces
The block lands on the plank and they move together. The system (block + plank) experiences a resistive force from the water given by $F = -kv$. We need to find the relation between the change in momentum and the distance traveled.
From Newton’s second law for the system of total mass $M_{total}$:
$$M_{total} \frac{dv}{dt} = -kv$$Integrating this with respect to time gives the impulse-momentum relation:
$$\int_{p_{initial}}^{p_{final}} dp = \int -kv \, dt = -k \int \frac{dx}{dt} \, dt = -k \int dx$$ $$\Delta p = -k \Delta x$$Step 2: Determine the required distance
The plank of length $l = 2.0 \text{ m}$ is initially at platform A. Platform A is at a distance $b = 17 \text{ m}$ from platform B. For the plank to “reach” platform B, its leading edge must touch B. Since the plank starts with its back at A (or close to it) and has length $l$, the distance the plank needs to travel to bridge the gap completely is the clear distance minus the plank’s length? Or simply the distance to touch B?
Let’s interpret the geometry standard for such problems: The gap between the platforms is $b$. The plank sits in the water. For the plank to just reach B (meaning establish a connection or touch it), the front of the plank must travel the distance equal to the gap minus the plank’s own length (assuming it starts fully adjacent to A). Therefore, the travel distance $\Delta x = b – l$.
$$\Delta x = 17 – 2 = 15 \text{ m}$$Step 3: Calculate the momentum change
Let $v_{block}$ be the initial speed of the block. When it lands on the plank (mass $M_p$), they acquire a common velocity $V_0$. By conservation of momentum during the landing:
$$p_{initial\_system} = m v_{block} = (m + M_p)V_0$$The system then decelerates due to water resistance until it stops ($v_f = 0$) just as it reaches platform B.
$$p_{final} – p_{initial\_system} = -k \Delta x$$ $$0 – m v_{block} = -k (b – l)$$ $$m v_{block} = k(b – l)$$Notice that the mass of the plank $M_p$ cancels out of the requirement for the initial impulse from the block.
Step 4: Solve for the minimum speed
Substitute the given values ($k = 15 \text{ Ns/m}$, $b = 17 \text{ m}$, $l = 2.0 \text{ m}$, $m = 25 \text{ kg}$):
$$v_{block} = \frac{k(b – l)}{m}$$ $$v_{block} = \frac{15(17 – 2)}{25}$$ $$v_{block} = \frac{15 \times 15}{25} = \frac{225}{25} = 9 \text{ m/s}$$