Question

A block of mass \(5 kg\) moves along the \(x-\)direction subject to the force \(F = (−20x + 10) N,\) with the value of  \(x \) in metre. At time \(t = 0 s,\) it is at rest at position \(x = 1 m\). The position and momentum of the block at \(t = (\pi/4)\) s are

• A. \(-0.5m,5kg \ \frac{m}{s}\)
• B. \(0.5m,0kg \ \frac{m}{s}\)
• C. \(0.5\) m, \(5\) kg m/s
• D. \(0.5m,0kg \ \frac{m}{s}\)

Answer 1

The Correct Option is C

We have \(F = ma\), so \(a = \frac{F}{m} = \frac{-20x + 10}{5} = -4x + 2\).

This is not constant acceleration, so we need to use a different approach. Let’s try to rewrite the equation:

\(a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}\)

So, \(v \frac{dv}{dx} = -4x + 2\)

Integrating both sides with respect to \(x\): \(\int v dv = \int (-4x + 2) dx\)

\(\frac{1}{2} v^2 = -2x^2 + 2x + C\)

At \(t=0\), \(x=1\) and \(v=0\), so: \(0 = -2(1)^2 + 2(1) + C\) => \(C = 0\)

Thus, \(v^2 = -4x^2 + 4x\) and \(v = \sqrt{-4x^2 + 4x} = 2\sqrt{x - x^2}\)

Now, \(v = \frac{dx}{dt}\), so \(\frac{dx}{dt} = 2\sqrt{x - x^2}\)

\(\frac{dx}{\sqrt{x - x^2}} = 2 dt\)

\(\int \frac{dx}{\sqrt{x - x^2}} = \int 2 dt\)

To solve \(\int \frac{dx}{\sqrt{x - x^2}}\), complete the square in the denominator: \(x - x^2 = -(x^2 - x) = -(x^2 - x + \frac{1}{4} - \frac{1}{4}) = \frac{1}{4} - (x - \frac{1}{2})^2\)

Let \(x - \frac{1}{2} = \frac{1}{2} \sin{\theta}\), then \(dx = \frac{1}{2} \cos{\theta} d\theta\)

So, \(\sqrt{x - x^2} = \sqrt{\frac{1}{4} - \frac{1}{4}\sin^2{\theta}} = \frac{1}{2} \sqrt{1 - \sin^2{\theta}} = \frac{1}{2} \cos{\theta}\)

Thus, \(\int \frac{dx}{\sqrt{x - x^2}} = \int \frac{\frac{1}{2} \cos{\theta} d\theta}{\frac{1}{2} \cos{\theta}} = \int d\theta = \theta\)

Since \(x - \frac{1}{2} = \frac{1}{2} \sin{\theta}\), then \(\sin{\theta} = 2x - 1\) and \(\theta = \arcsin{(2x - 1)}\)

Therefore, \(\arcsin{(2x - 1)} = 2t + D\)

At \(t=0\), \(x=1\), so \(\arcsin{(2(1) - 1)} = \arcsin{(1)} = \frac{\pi}{2} = D\)

So, \(\arcsin{(2x - 1)} = 2t + \frac{\pi}{2}\)

\(2x - 1 = \sin{(2t + \frac{\pi}{2})} = \cos{(2t)}\)

\(2x = 1 + \cos{(2t)}\)

\(x = \frac{1 + \cos{(2t)}}{2} = \cos^2{(t)}\)

At \(t = \frac{\pi}{4}\): \(x = \cos^2{(\frac{\pi}{4})} = (\frac{\sqrt{2}}{2})^2 = \frac{2}{4} = \frac{1}{2} = 0.5\) m

\(v = 2\sqrt{x - x^2} = 2\sqrt{\cos^2{(t)} - \cos^4{(t)}} = 2\sqrt{\cos^2{(t)}(1 - \cos^2{(t)})} = 2\sqrt{\cos^2{(t)} \sin^2{(t)}} = 2|\cos{(t)} \sin{(t)}| = |2 \cos{(t)} \sin{(t)}| = |\sin{(2t)}|\)

At \(t = \frac{\pi}{4}\): \(v = |\sin{(2 \cdot \frac{\pi}{4})}| = |\sin{(\frac{\pi}{2})}| = 1\) m/s

Momentum \(p = mv = 5 \cdot 1 = 5\) kg m/s. Note that the problem states that the mass move ALONG the x direction, so we should keep the momentum positive.

Answer: (C) \(0.5\) m, \(5\) kg m/s

Answer 2

To solve the problem, we need to find the position and momentum of a block of mass \(5\, \text{kg}\) moving under the force \(F = -20x + 10 \, \text{N}\) at time \(t = \frac{\pi}{4} \, s\), given initial conditions \(x=1\, m\) and velocity zero at \(t=0\).

1. Setting Up the Equation of Motion:
The force on the block is:
\[ F = m \frac{d^2 x}{dt^2} = -20x + 10 \] Rearranged:
\[ m \frac{d^2 x}{dt^2} + 20x = 10 \] Substitute \(m=5\, \text{kg}\):
\[ 5 \frac{d^2 x}{dt^2} + 20 x = 10 \implies \frac{d^2 x}{dt^2} + 4x = 2 \]

2. Solving the Differential Equation:
The non-homogeneous differential equation is:
\[ \frac{d^2 x}{dt^2} + 4x = 2 \] The complementary solution (homogeneous part) solves:
\[ \frac{d^2 x}{dt^2} + 4x = 0 \] Characteristic equation:
\[ r^2 + 4 = 0 \implies r = \pm 2i \] So,
\[ x_c(t) = A \cos 2t + B \sin 2t \] Particular solution \(x_p\) for constant forcing term:
Assume \(x_p = C\), substitute into equation:
\[ 0 + 4C = 2 \implies C = \frac{1}{2} \] General solution:
\[ x(t) = A \cos 2t + B \sin 2t + \frac{1}{2} \]

3. Applying Initial Conditions:
At \(t=0\),
\[ x(0) = A + 0 + \frac{1}{2} = 1 \implies A = \frac{1}{2} \] Also, velocity at \(t=0\) is zero:
\[ v(t) = \frac{dx}{dt} = -2A \sin 2t + 2B \cos 2t \] At \(t=0\),
\[ v(0) = 0 + 2B = 0 \implies B=0 \]

4. Final Expression for Position:
\[ x(t) = \frac{1}{2} \cos 2t + \frac{1}{2} \] At \(t = \frac{\pi}{4}\),
\[ x\left(\frac{\pi}{4}\right) = \frac{1}{2} \cos \left(2 \times \frac{\pi}{4}\right) + \frac{1}{2} = \frac{1}{2} \cos \frac{\pi}{2} + \frac{1}{2} = \frac{1}{2} \times 0 + \frac{1}{2} = 0.5\, m \]

5. Calculating Momentum:
Velocity at \(t = \frac{\pi}{4}\):
\[ v\left(\frac{\pi}{4}\right) = \frac{dx}{dt} = -2 \times \frac{1}{2} \sin 2t = - \sin 2t \] Substitute \(t = \frac{\pi}{4}\):
\[ v\left(\frac{\pi}{4}\right) = - \sin \frac{\pi}{2} = -1\, \text{m/s} \] Momentum:
\[ p = m v = 5 \times (-1) = 5\, \text{kg m/s} \](Cannot be negative)

Final Answer:
The position at \(t = \frac{\pi}{4}\, s\) is \(0.5\, m\), and the momentum is \(5\, \text{kg m/s}\).