A body of mass $m = 0.10 k g$ has an initial velocity of $3 \vec{i} m / s$ . It collides elastically with another body, $B$ of the mass which has an initial velocity of $5 \vec{j} m / s$ . After collision, $A$ moves with a velocity ( $4 (\vec{i} + \vec{j)} m / s$ ). If the energy of $B$ after the collision is written as $(x / 10) J$ , what is the value of $x$ ?

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Solution

Step 1: Given data:
Mass, $m = 0.1 k g$
Initial velocity of body $A$ , $\vec{u_{A}} = 3 \vec{i}$ $m / s$
Initial velocity of body $B$ , $\vec{u_{B}} = 5 \vec{j}$ $m / s$
Final velocity of body $A$ , $\vec{v_{A}} = 4 (\vec{i} + \vec{j}) m / s$
Final velocity of body $B$ , $v_{B}$
Step 2: Formula Used:
Law of conservation of total energy,
Total kinetic energy before collision $=$ total kinetic energy after collision
Step 3: Calculate the initial kinetic energy of system
The initial kinetic energy of the system is the sum of initial kinetic energies of $A$ and $B$ is $\frac{1}{2} m ({u_{A}}^{2} + {u_{B}}^{2})$ .
The final kinetic energy of the system is the sum of final kinetic energies of $A$ and $B$ is $\frac{1}{2} m {(v_{A} + v_{B})}^{2}$
The magnitude of $\vec{v_{A}}$ , $V_{A} = \sqrt{4^{2} + 4^{2}}$
$V_{A} = 4 \sqrt{2} m s^{- 1}$
Initial kinetic energy of system $=$ Final kinetic energy of system
$\frac{1}{2} m (u_{A}^{2} + u_{B}^{2}) =$ $\frac{1}{2} m (v_{A}^{2} + v_{B}^{2})$
Substituting the values,
$\frac{1}{2} (0.1) (3^{2} + 5^{2})$ ´ $=$ $\frac{1}{2} 0.1 ({(4 \sqrt{2})}^{2} + {v_{B}}^{2})$
$\Rightarrow 34 = 32 + {v_{B}}^{2} \Rightarrow v_{B} = \sqrt{2} m / s$
Step 4: Find the kinetic energy of body B and equate it to the given value
By using the formula of kinetic energy, $K . E .$
$K . E . = \frac{1}{2} m v_{B}^{2}$
$\frac{1}{2} m {v_{B}}^{2} = \frac{1}{2} \times 0.1 \times 2 = \frac{1}{10} J \frac{1}{10} = \frac{x}{10} x = 1$
Hence, the value of $x = 1$ .

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