Question

A body A, of mass $m=0$.$1kg$ has an initial velocity of $3\hat{i}m/s$ . It collides elastically with another body, $B$ of the same mass which has an initial velocity of $5\hat{j}m/s$. After collision, A moves with a velocity $\overrightarrow{v}=4(\hat{i}+\hat{j})$ The energy of $B$ after collision is written as $\frac{x}{10}J$. The value of $x$ is

Answer 1

Step 1: Find the velocity of the body $B$
let $u_1=3\hat{i}m/s,m_1=m_2=0.1kg,u_2=5\hat{j}m/s,v_1=\overrightarrow{v}=4(\hat{i}+\hat{j})$
By conservation of linear momentum:
$\left(0.1\right)\left(3\hat{i}\right)+\left(0.1\right)\left(5\hat{j}\right)=\left(0.1\right)\left(4\right)(\hat{i}+\hat{j})+\left(0.1\right)v_B$
$v_B=-\hat{i}+\hat{j}$

Step 2: Find the kinetic energy of the body $B$
$\therefore$Speed of $B$ after collision$\left|v_B\right|=\sqrt{2}m/s$
Now, kinetic energy of$B$
$K_B=\frac{1}{2}m{v}^2=\frac{1}{2}\left(0.1\right)\left(2\right)=\frac{1}{10}J$
$\therefore x=1$
Final answer: $1$