Question

A body A, of mass $m = 0.1$ kg has an initial velocity of $3\hat{i}\,ms^{-1}$. It collides elastically with another body, B of the same mass which has an initial velocity of $5\hat{j}\,ms^{-1}$ After collision, A moves with a velocity $\vec{\nu} = 4\left(\hat{i}+\hat{j}\right)$. The energy of B after collision is written as $\frac{x}{10} J$. The value of x is __________ .

Answer 1

By conservation of linear momentum :
$\left(0.1\right)\left(3\hat{i}\right)+\left(0.1\right)\left(5\hat{j}\right)=\left(0.1\right)\left(4\right)\left(\hat{i}+\hat{j}\right)+\left(0.1\right)\vec{v}$
$\Rightarrow \vec{v}=-\hat{i}+\hat{j}$
$\therefore $ Speed of B after collision $\left|\vec{v}\right|=\sqrt{2}$
Now, kinetic energy $=\frac{1}{2}mV^{2}=\frac{1}{2}\left(0.1\right)\left(2\right)=\frac{1}{10}$
$\therefore x=1$