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Step-by-Step Solution
Step 1: Identify the given information
• Mass of body A, m_A = 0.1 \text{ kg} , initial velocity of A, \vec{u}_A = 3\,\hat{i}\,\text{m s}^{-1} .
• Mass of body B, m_B = 0.1 \text{ kg} , initial velocity of B, \vec{u}_B = 5\,\hat{j}\,\text{m s}^{-1} .
• After collision, velocity of A, \vec{v}_A = 4(\hat{i} + \hat{j}) .
• We want to find the kinetic energy of B after collision, which is given in the form \frac{x}{10} , and hence find the value of x .
Step 2: Use the law of conservation of linear momentum
Because the collision is elastic (and no external forces act in the horizontal plane), total momentum before collision equals total momentum after collision:
\displaystyle m_A \vec{u}_A + m_B \vec{u}_B = m_A \vec{v}_A + m_B \vec{v}_B\,.
Substitute the given values:
\displaystyle (0.1)(3\,\hat{i}) + (0.1)(5\,\hat{j})
= (0.1)\bigl[4(\hat{i} + \hat{j})\bigr] + (0.1)\vec{v}_B\,.
Step 3: Simplify to find the velocity of body B
First, compute the left-hand side (LHS) momenta numerically:
\displaystyle 0.1 \times 3\,\hat{i} = 0.3\,\hat{i}, \quad 0.1 \times 5\,\hat{j} = 0.5\,\hat{j}.
So, LHS = 0.3\,\hat{i} + 0.5\,\hat{j}.
On the right-hand side (RHS):
\displaystyle (0.1)\bigl[4(\hat{i} + \hat{j})\bigr] = 0.4\,\hat{i} + 0.4\,\hat{j}, \quad (0.1)\vec{v}_B = 0.1\,\vec{v}_B.
Hence the momentum equation is:
\displaystyle 0.3\,\hat{i} + 0.5\,\hat{j} = 0.4\,\hat{i} + 0.4\,\hat{j} + 0.1\,\vec{v}_B.
Rearrange to solve for \vec{v}_B :
\displaystyle 0.1\,\vec{v}_B = (0.3\,\hat{i} - 0.4\,\hat{i}) + (0.5\,\hat{j} - 0.4\,\hat{j}) = -0.1\,\hat{i} + 0.1\,\hat{j}.
\displaystyle \vec{v}_B = -\hat{i} + \hat{j}.
Step 4: Calculate the speed of body B after collision
The velocity of B is \vec{v}_B = -\hat{i} + \hat{j} . Its magnitude is:
\displaystyle \left|\vec{v}_B\right| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}\,.
Step 5: Determine the kinetic energy of body B
Kinetic energy (K.E.) of B after collision is:
\displaystyle \text{K.E.}_B = \frac{1}{2} \, m_B \, \left|\vec{v}_B\right|^2.
Substitute m_B = 0.1\,\text{kg} and \left|\vec{v}_B\right| = \sqrt{2} :
\displaystyle \text{K.E.}_B
= \frac{1}{2} \times 0.1 \times (\sqrt{2})^2
= \frac{1}{2} \times 0.1 \times 2
= 0.1 \,\text{J}
= \frac{1}{10}\,\text{J}.
Step 6: Identify the value of x
Since the kinetic energy is written as \displaystyle \frac{x}{10}\,\text{J} , and we found it to be \displaystyle \frac{1}{10}\,\text{J} , it follows that
\displaystyle x = 1.
