COM BYU 4

Step 1: Initial State Analysis

At time $t=0$, the car (mass $m$) starts moving with a constant velocity $v$ relative to the plank (mass $M$). The plank is on a horizontal floor with friction coefficient $\mu$. Initially, the system is at rest. Let $v_c$ be the car’s velocity and $v_p$ be the plank’s velocity with respect to the ground.

From the relative velocity condition:

$$v_c – v_p = v \implies v_c = v_p + v$$

Using conservation of momentum just after the start (before external friction acts significantly):

$$m v_c + M v_p = 0$$ $$m(v_p + v) + M v_p = 0$$ $$v_p (m + M) = -mv \implies v_p = -\frac{mv}{m+M}$$

The plank moves to the left initially, and the car moves to the right with initial velocity:

$$v_{c,initial} = v_p + v = -\frac{mv}{m+M} + v = \frac{Mv}{m+M}$$

Step 2: Dynamics and Time Calculation

The plank moves left, so kinetic friction $f_k$ acts on it to the right. The same friction force accelerates the car-plank system to the right.

$$f_k = \mu N = \mu (m+M)g$$

Acceleration of the system ($a$):

$$a = \frac{f_k}{m+M} = \mu g$$

The plank decelerates and eventually stops. The time $t_0$ taken for the plank to stop is:

$$0 = v_{p,initial} + a t_0$$ $$0 = -\frac{mv}{m+M} + \mu g t_0$$ $$t_0 = \frac{mv}{\mu g(m+M)}$$

Step 3: Velocity as a Function of Time

Case 1: $0 \le t \le t_0$ (Plank is sliding)
The car accelerates with $a = \mu g$. Its velocity is:

$$v_c(t) = v_{c,initial} + a t = \frac{Mv}{m+M} + \mu g t$$

Case 2: $t > t_0$ (Plank stops)
Once the plank stops, the friction becomes static or zero (since the car moves at constant relative velocity $v$ and the plank is stationary, the car simply moves at $v$ relative to the ground). The velocity remains constant.

$$v_c(t) = v_{plank} + v = 0 + v = v$$