Question

A force on an object of mass \(100\ g\) is \((10\hat i+5\hat j)\ N\). The position of that object at \(t = 2s\) is \((a\hat i+b\hat j)\ m\) after starting from rest. The value of \(\frac ab\) will be _______ .

Answer 1

Correct Answer: 2

 Given the net force \(\vec{F} = (10\hat{i} + 5\hat{j})\ N\) on an object of mass \(m = 100\ g = 0.1\ kg\), we need to find the position at \(t = 2\ s\) after starting from rest and calculate \(\frac{a}{b}\) where the position is \((a\hat{i} + b\hat{j})\ m\).

1. **Calculate Acceleration:**

Using Newton's second law: \(\vec{F} = m\vec{a}\), we have:

\(\vec{a} = \frac{\vec{F}}{m} = \frac{10 \hat{i} + 5 \hat{j}}{0.1} = 100\hat{i} + 50\hat{j}\ m/s^2\)

2. **Find Velocity at \(t = 2\ s\):**

The initial velocity \(\vec{u} = 0\) (at rest), use \(\vec{v} = \vec{u} + \vec{a}t\):

\(\vec{v} = 0 + (100\hat{i} + 50\hat{j}) \times 2 = 200\hat{i} + 100\hat{j}\ m/s\)

3. **Calculate Position at \(t = 2\ s\):**

Using the kinematic equation: \(\vec{s} = \vec{u}t + \frac{1}{2}\vec{a}t^2\):

\(\vec{s} = 0 + \frac{1}{2}(100\hat{i} + 50\hat{j})(2^2) = 200\hat{i} + 100\hat{j}\ m\)

Thus, \(a = 200\) and \(b = 100\).

4. **Calculate \(\frac{a}{b}\):\)

\(\frac{a}{b} = \frac{200}{100} = 2\)

5. **Validation:**

The calculated \(\frac{a}{b} = 2\) falls within the provided range (2, 2).