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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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The correct answer is:
\(-\frac{50}{3} \mathrm{~ms}^{-1}, \frac{10}{3} \mathrm{~ms}^{-1}\)
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
\(\begin{aligned}
& \frac{10}{3}-v_{1 f}=20 \\
& \Rightarrow \quad-v_{1 f}=20-\frac{10}{3}=\frac{50}{3} \\
& \Rightarrow \quad v_{1 f}=\frac{-50}{3} \mathrm{~ms}^{-1} \\
\end{aligned}\)
\(v_1=v_2=10 \mathrm{~ms}^{-1}\)
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
\(\begin{aligned}
& \frac{10}{3}-v_{1 f}=20 \\
& \Rightarrow \quad-v_{1 f}=20-\frac{10}{3}=\frac{50}{3} \\
& \Rightarrow \quad v_{1 f}=\frac{-50}{3} \mathrm{~ms}^{-1} \\
\end{aligned}\)
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