Two point-particles having masses \(m_1\) and \(m_2\) approach each other in perpendicular directions with speeds \(v_1\) and \(v_2\), respectively, as shown in the figure below. After an elastic collision, they move away from each other in perpendicular directions with speeds \(v'_1\) and \(v'_2\), respectively.
The ratio \(\frac{v'_1}{v_1}\) is
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In collision problems, always start with the fundamental conservation laws of momentum and energy. Special geometric conditions, like perpendicular velocities, can greatly simplify the algebra by making dot products zero. If you arrive at a result that contradicts the options, double-check your derivation, but also be aware that questions in exam papers can sometimes be flawed.
Step 1: Understanding the Concept:
This problem involves a two-dimensional elastic collision. For any collision, linear momentum is conserved. For an elastic collision, kinetic energy is also conserved. The problem states that the initial velocities are perpendicular to each other, and the final velocities are also perpendicular to each other. Step 2: Key Formula or Approach:
1. Conservation of Linear Momentum (Vector): \(\mathbf{P}_{\text{initial}} = \mathbf{P}_{\text{final}}\) \(\implies\) \(m_1\mathbf{v}_1 + m_2\mathbf{v}_2 = m_1\mathbf{v}'_1 + m_2\mathbf{v}'_2\).
2. Conservation of Kinetic Energy (Scalar): \(K_{\text{initial}} = K_{\text{final}}\) \(\implies\) \(\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2}m_1 (v'_1)^2 + \frac{1}{2}m_2 (v'_2)^2\).
3. Perpendicularity Conditions: \(\mathbf{v}_1 \cdot \mathbf{v}_2 = 0\) and \(\mathbf{v}'_1 \cdot \mathbf{v}'_2 = 0\). Step 3: Detailed Explanation:
From the conservation of linear momentum, we can square the vector equation:
\[ (m_1\mathbf{v}_1 + m_2\mathbf{v}_2) \cdot (m_1\mathbf{v}_1 + m_2\mathbf{v}_2) = (m_1\mathbf{v}'_1 + m_2\mathbf{v}'_2) \cdot (m_1\mathbf{v}'_1 + m_2\mathbf{v}'_2) \]
Expanding this using the distributive property of the dot product:
\[ m_1^2 v_1^2 + m_2^2 v_2^2 + 2m_1m_2(\mathbf{v}_1 \cdot \mathbf{v}_2) = m_1^2 (v'_1)^2 + m_2^2 (v'_2)^2 + 2m_1m_2(\mathbf{v}'_1 \cdot \mathbf{v}'_2) \]
Using the perpendicularity conditions (\(\mathbf{v}_1 \cdot \mathbf{v}_2 = 0\) and \(\mathbf{v}'_1 \cdot \mathbf{v}'_2 = 0\)), this simplifies to:
\[ m_1^2 v_1^2 + m_2^2 v_2^2 = m_1^2 (v'_1)^2 + m_2^2 (v'_2)^2 \quad \text{(Equation 1)} \]
The conservation of kinetic energy equation is:
\[ m_1 v_1^2 + m_2 v_2^2 = m_1 (v'_1)^2 + m_2 (v'_2)^2 \quad \text{(Equation 2)} \]
We now have a system of two equations. Let's rearrange Equation 2:
\[ m_1(v_1^2 - (v'_1)^2) = m_2((v'_2)^2 - v_2^2) \quad \text{(Equation 3)} \]
And rearrange Equation 1:
\[ m_1^2(v_1^2 - (v'_1)^2) = m_2^2((v'_2)^2 - v_2^2) \quad \text{(Equation 4)} \]
Let \(X = v_1^2 - (v'_1)^2\) and \(Y = (v'_2)^2 - v_2^2\). The equations become:
\(m_1 X = m_2 Y\) and \(m_1^2 X = m_2^2 Y\).
Substituting \(Y = \frac{m_1}{m_2}X\) into the second equation gives:
\[ m_1^2 X = m_2^2 \left(\frac{m_1}{m_2}X\right) \implies m_1^2 X = m_1 m_2 X \]
\[ (m_1^2 - m_1 m_2)X = 0 \implies m_1(m_1 - m_2)X = 0 \]
This implies that either \(m_1=m_2\) or \(X=0\).
If \(X = v_1^2 - (v'_1)^2 = 0\), then \(v_1 = v'_1\).
