Displacement of Block on Oily Surface Under Exponential Velocity

An impulse $J$ delivered to a block of mass $m$ initially at rest imparts an initial velocity $v(0^+) = v_0$. According to the impulse-momentum theorem: $J = \Delta p = m v_0 - 0$

Given:

• Mass of the block, $m = 0.4 \text{ kg}$
• Magnitude of impulse, $J = 1.0 \text{ N s}$
• Time constant, $\tau = 4 \text{ s}$
• $e^{-1} = 0.37$

We can find the initial velocity $v_0$: $v_0 = \frac{J}{m} = \frac{1.0 \text{ N s}}{0.4 \text{ kg}} = 2.5 \text{ m/s}$

The velocity of the block as a function of time $t$ is given by: $v(t) = v_0 e^{-t/\tau}$

The displacement $x(t)$ of the block at time $t = \tau$ is obtained by integrating the velocity with respect to time from $t = 0$ to $t = \tau$: $x(\tau) = \int_{0}^{\tau} v(t) \, dt = \int_{0}^{\tau} v_0 e^{-t/\tau} \, dt$

Evaluating the integral: $x(\tau) = v_0 \left[ -\tau e^{-t/\tau} \right]_{0}^{\tau} = v_0 \tau \left( 1 - e^{-1} \right)$

Substituting the given and calculated values into the expression: $x(\tau) = (2.5 \text{ m/s}) \times (4 \text{ s}) \times (1 - 0.37)$ $x(\tau) = 10 \times 0.63 = 6.3 \text{ m}$

Thus, the displacement of the block at $t = \tau$ is $6.30$ (or $6.3$) metres.