Question

A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass \(m=0.4 \mathrm{~kg}\) is at rest on this surface. An impulse of \(1.0 \mathrm{~N} \mathrm{~s}\) is applied to the block at time \(t=0\) so that it starts moving along the \(x\) -axis with a velocity \(v(t)=v_{0} e^{-t / \tau}\), where \(v_{0}\) is a constant and \(\tau=4 \mathrm{~s}\). The displacement of the block, in metres, at \(t=\tau\) is Take \(e^{-1}=0.37\).

Short answer

The displacement of the block at time \( t = \tau \) is 6.3 meters.

Step-by-step solution

Apply Impulse-Momentum Theorem

The impulse-momentum theorem states that the change in momentum of an object is equal to the impulse applied to it. This can be written as \( J = \Delta p = m(v_f - v_i) \). The initial speed \( v_i \) is 0 since the block starts from rest. We are given the impulse \( J = 1.0 \, \text{N}\cdot\text{s} \) and the mass of the block \( m = 0.4 \, \text{kg} \). Hence, the initial velocity \( v_0 \) can be found by \( v_0 = \frac{J}{m} \).

Calculate the Initial Velocity

Using the impulse-momentum theorem, calculate the initial velocity: \( v_0 = \frac{1.0 \, \text{N}\cdot\text{s}}{0.4 \, \text{kg}} = \frac{1}{0.4} = 2.5 \, \text{m/s} \). This is the initial velocity \( v_0 \) of the block right after the impulse has been applied.

Understand the Velocity Function

The block starts moving with a velocity that changes over time according to the velocity function \( v(t) = v_0 e^{-t / \tau} \), where \( \tau = 4 \, \text{s} \) is the time constant. This function accounts for the deceleration of the block over time due to the thin layer of oil.

Determine Displacement at Time \( t = \tau \)

To find the displacement \( s(t) \), integrate the velocity function \( v(t) \) with respect to time. The displacement at any time \( t \) is given by \( s(t) = \int v(t) \, dt = \int v_0 e^{-t/\tau} \, dt \). Since the limit of integration is from 0 to \( \tau \), the displacement \( s(\tau) \) at time \( t = \tau \) is the integral evaluated over this interval.

Evaluate the Integral

Integrating the velocity function we get \( s(\tau) = \int_{0}^{\tau} v_0 e^{-t/\tau} \, dt \). This integral can be solved by recognizing it as a definite integral of an exponential function. The result is \( s(\tau) = -v_0 \tau e^{-t/\tau} |_{0}^{\tau} \), which can be further simplified using the provided value of \( e^{-1} = 0.37 \).

Calculate the Displacement at \( t = \tau \)

Plugging in the limits and simplifying the expression for displacement we get \( s(\tau) = -v_0 \tau (e^{-\tau/\tau} - e^{0}) = -v_0 \tau (0.37 - 1) = -2.5 \times 4 \times (-0.63) \).

Find the Numerical Value of Displacement

Calculate the numerical value of displacement \( s(\tau) = -2.5 \times 4 \times (-0.63) = 6.3 \, \text{m} \). Since displacement cannot be negative, the negative signs cancel each other, and the block has traveled 6.3 meters by the time \( t = \tau \).

Key concepts

Kinematics

In the context of physics, kinematics is the study of motion without considering the forces that cause that motion. It deals with concepts such as velocity, acceleration, displacement, and time.

For the given exercise, kinematics is essential to understand how the velocity of the block changes over time and how this velocity influences the displacement of the block. After an impulse is applied, the block starts moving with an initial velocity, which gradually decreases due to the frictional force of the oil layer - a key aspect of the block's kinematic description.

Students often struggle with the concept that displacement is not just the product of velocity and time, especially when velocity is not constant. The exercise requires us to use integration of the velocity over time to find the displacement accurately, rather than just multiplying the initial velocity by time.

Exponential Decay

Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value. This concept is commonly found in physics, chemistry, biology, and finance, among other fields. In the case of our exercise, the velocity of the block decays exponentially because the frictional force acts on the block in the opposite direction of its motion.

The general form of an exponential decay function can be written as \( A(t) = A_0 e^{-\frac{t}{\tau}} \), where \( A_0 \) is the initial value, \( \tau \) is the decay constant, and \( e \) is the base of the natural logarithm. Understanding this idea helps students visualize how the velocity of the block changes over time, decreasing continuously rather than at a constant rate, and is critical for understanding the motion described in the exercise.

Definite Integral

The definite integral is a fundamental concept in calculus that is used to find the accumulation of a quantity, such as area under a curve, total displacement, or any other quantity that accumulates over a certain interval. In our problem, we used the definite integral to find the total displacement of the block over the time interval from 0 to \( \tau \).

To understand this, consider the velocity function \( v(t) = v_0 e^{-\frac{t}{\tau}} \), which gives us the rate of change of displacement at any particular time. By integrating this function over a range of time, we accumulate all the small changes in displacement to find the total displacement over that period. It's important to recognize that a definite integral has limits – in this case, from 0 to \( \tau \) – which define the interval over which we are accumulating our quantity. Students sometimes confuse this with an indefinite integral, which lacks these bounds and includes a constant of integration, representing a family of functions rather than a single value.