\(m_{A}=300 g m_{B}=900 g \quad\) After collision
\(=0.3 kg =0.9 kg\)
\(u_{A}=6 m / s \quad u_{B}=0\)efore collision
Note:
We notice from the problem that the collision was elastic which implies that K.E. and linear momentum are conserved. By the law of conservation of linear momentum; total linear momentum before collision = total linear momentum after collision i.e.
$$
\begin{aligned}
&m_{A} u_{A}+m_{B} \pi_{B}=m_{A} v_{A}+m_{B} v_{B} \\
& M _{ A } U _{ A }= M _{ A } V _{ A }+ M _{ B } V _{ B } \\
&0.3 \times 6=0.3 \times V _{ A }+0.9 \times V _{ B } \\
&1.8=0.3 V _{ A }+0.9 V _{ B } \ldots \ldots \ldots \ldots \ldots \text { (1) } \\
&\text { Equation (1) has two...... }
\end{aligned}
$$quation (1) has two unknowns, it cannot be solved. So. we seck another equation ir terms of \(V_{A}\) and \(V_{B}\). The second equation is conservat in of kinetic energy i.e. total K.E energy before collis \({ }^{\circ}\) on \(=\) total K.E. after collisior thus;
\begin{aligned}
&\frac{1}{2} m_{A} u_{A}^{2}+\frac{1}{2} n_{B} u_{B}^{2}=\frac{1}{2} m_{A} v_{A}^{2}+\frac{1}{2} m_{B} v_{B}^{2} \\
&\frac{1}{2} m_{A} u_{A}^{2}=\frac{1}{2} m_{A} v_{A}^{2}+\frac{1}{2} m_{B} v_{B}^{2} \\
&m_{A} u_{A}^{2}=m_{A} v_{A}^{2}+m_{B} v_{B}^{2} \\
&0.3 \times 6^{2}=0.3 \times v_{A}^{2}+0.9 \times v_{B}^{2} \\
&10.8=0.3 v_{A}^{2}+0.9 v_{B}^{2} \\
&\text { Divide through by } 0.3 \\
&36=v_{A}^{2}+3 v_{B}^{2} \cdots \ldots \ldots \ldots(2) \\
&\text { from eqn }(1), \\
&1.8=0.3 V_{A}+0.9 V_{B} \\
&\text { divide through by } 0.3 \\
&6=V_{A}+3 V_{B} \\
&\text { Or } V_{A}=6-3 V_{B} \ldots \ldots \ldots \ldots . . . \\
&\text { put }(3) \text { into }(2) \\
&36=\left(6-3 V_{B}\right)^{2}+3 v_{B}^{2} \\
&36=36-36 V_{B}+9 v_{B}^{2}+3 v_{B}^{2} \\
&0=-36 V_{B}+12 v_{B}^{2} \\
&\text { Or } \\
&12 v_{B}^{2}-36 V_{B}=0 \\
&V_{B}\left(12 V_{B}-36\right)=0 \\
&V_{B}=0 \text { or } 12 V_{B}-36=0 \\
&12 V_{B}=36
\end{aligned}
$$
\begin{aligned}
&V_{B}=\frac{36}{12} \\
&V_{B}=3 m / s \\
&\text { from }(3) \\
&V_{A}=6-3 V_{B} \\
&V_{A}=6-3(3) \\
&V_{A}=6-9=-3 m / s
\end{aligned}
$$
this result implies that the heavy mass moves forward with a speed of \(3 m / s\) while the lighter mass moves backward with a speed of \(3 m / s\)