A body A of mass $m = 0.1$ kg has an initial velocity of $3\hat i\ \text{m s}^{-1}$. It collides elastically with another body B of the same mass which has an initial velocity of $5\hat j\ \text{m s}^{-1}$. After the collision, A moves with a velocity $\vec v = 4(\hat i + \hat j)\ \text{m s}^{-1}$. The energy of B after the collision is written as $\dfrac{x}{10}$ J. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 1
Momentum conservation (equal masses): $3\hat i + 5\hat j = 4\hat i + 4\hat j + \vec v_B$, so $\vec v_B = -\hat i + \hat j$.
$$K_B = \frac12\times0.1\times(1 + 1) = 0.1\ \text{J} = \frac{1}{10}\ \text{J}$$
(Check: total KE before $= 0.05\times34 = 1.7$ J, after $= 0.05\times(32 + 2) = 1.7$ J.)
So $x = 1$.
Solution by Sreeraj P, M.Sc Physics