Two balls $X(2 \mathrm{~kg})$ and $Y(4 \mathrm{~kg})$ approach each other with equal speeds of $10 \mathrm{~ms}^{-1}$. If the collision is perfectly elastic, then the new velocities of balls $X$ and $Y$ are respectively

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Aicharya · Accepted Answer

Given, $m_1=2 \mathrm{~kg}, m_2=4 \mathrm{~kg}$ $v_1=v_2=10 \mathrm{~ms}^{-1}$ In perfectly elastic collision, momentum and kinetic energy is conserved. $\therefore$ By conservation of momentum, $\begin{aligned} & m_1 \times 10+m_2 \times(-10)=m_1 v_{1 f}+m_2 v_{2 f} \ & \Rightarrow \quad 2 \times 10+4 \times(-10)=2 v_{1 f}+4 v_{2 f} \ & \Rightarrow \quad-20=2\left(v_{1 f}+2 v_{2 f}\right) \ & \Rightarrow \quad v_{1 f}+2 v_{2 f}=-10 \quad \ldots (i) \ \end{aligned}$ In perfectly elastic collision, Velocity of separation $=$ Velocity of approach $\begin{array}{ll} \Rightarrow & v_{2 f}-v_{1 f}=10-(-10) \ \Rightarrow & v_{2 f}-v_{1 f}=20 \quad \ldots (ii) \end{array}$ Adding Eqs. (i) and (ii), we get $\begin{aligned} & 3 v_{2 f}=10 \ & v_{2 f}=\frac{10}{3} \mathrm{~ms}^{-1} \end{aligned}$ From Eqs. (ii), we get $\begin{aligned} & \frac{10}{3}-v_{1 f}=20 \ & \Rightarrow \quad-v_{1 f}=20-\frac{10}{3}=\frac{50}{3} \ & \Rightarrow \quad v_{1 f}=\frac{-50}{3} \mathrm{~ms}^{-1} \ \end{aligned}$

Two balls \(X(2 \mathrm{~kg})\) and \(Y(4 \mathrm{~kg})\) approach each other with equal speeds of \(10…

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