Question:
Consider two particles, each of mass 20 g; the first particle is moving with a speed of 10 m/s along a one-dimensional track in the positive x-direction and collides with the second particle at rest. Assuming that the collision is elastic, the speed (in m/s) of the first particle after the collision is ...............
Consider two particles, each of mass 20 g; the first particle is moving with a speed of 10 m/s along a one-dimensional track in the positive x-direction and collides with the second particle at rest. Assuming that the collision is elastic, the speed (in m/s) of the first particle after the collision is ...............
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In elastic collisions with equal masses, the velocities of the particles are swapped.
Updated On: Dec 2, 2025
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Solution and Explanation
Step 1: Conservation of momentum.
Since the collision is elastic, both momentum and kinetic energy are conserved. The general equations for an elastic collision are:
\[
m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2,
\]
where \( m_1 \) and \( m_2 \) are the masses of the particles, \( u_1 \) and \( u_2 \) are the initial velocities, and \( v_1 \) and \( v_2 \) are the final velocities of the particles.
For a perfectly elastic collision of equal masses:
\[
v_1 = u_2 \text{and} v_2 = u_1.
\]
Step 2: Applying the known values.
We are given \( u_1 = 10 \, \text{m/s} \), \( u_2 = 0 \, \text{m/s} \), and the masses are equal (\( m_1 = m_2 = 20 \, \text{g} \)).
So the final speed of the first particle after the collision is:
\[
v_1 = u_2 = 0 \, \text{m/s}.
\]
Step 3: Conclusion.
The final speed of the first particle after the collision is \( \boxed{0} \, \text{m/s} \).
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