Q.45 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 ________.
The speed of the first particle after collision is 0 m/s
Problem Q.45: Two particles, each of mass 20 g; first particle moving with speed 10 m/s along positive x-direction collides elastically with second particle at rest. Find speed of first particle after collision.
Problem Setup
Two particles each have mass \( m_1 = m_2 = 0.02 \) kg. Initial velocities: \( u_1 = 10 \) m/s (positive x-direction), \( u_2 = 0 \) m/s. Collision is elastic, conserving momentum and kinetic energy.
Conservation Equations
Momentum Conservation:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
\[ 0.02 \times 10 + 0.02 \times 0 = 0.02 v_1 + 0.02 v_2 \]
\[ \boxed{10 = v_1 + v_2} \quad (1) \]
Kinetic Energy Conservation:
\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
\[ 0.02 \times 10^2 = 0.02 (v_1^2 + v_2^2) \]
\[ \boxed{100 = v_1^2 + v_2^2} \quad (2) \]
Solution Derivation
From equation (1): \( v_2 = 10 – v_1 \)
Substitute into equation (2):
\[ 100 = v_1^2 + (10 – v_1)^2 \]
\[ 100 = v_1^2 + 100 – 20 v_1 + v_1^2 \]
\[ 0 = 2 v_1^2 – 20 v_1 \]
\[ v_1 (v_1 – 10) = 0 \]
\[ \boxed{v_1 = 0 \, \text{m/s} \quad \text{or} \quad v_1 = 10 \, \text{m/s}} \]
Conclusion: \( v_1 = 10 \) m/s implies no collision. Valid solution is \( v_1 = 0 \) m/s, \( v_2 = 10 \) m/s.
Special Case Insight
General Formula (when \( u_2 = 0 \)):
\[ v_1 = \frac{m_1 – m_2}{m_1 + m_2} u_1 \]
\[ v_1 = \frac{0.02 – 0.02}{0.02 + 0.02} \times 10 = 0 \, \text{m/s} \]
Key Physics Principle: In 1D elastic collisions with equal masses and one at rest, velocities are exchanged completely.
Verification by Code
m1 = 0.02; m2 = 0.02; u1 = 10; u2 = 0
v1 = ((m1 - m2)/(m1 + m2)) * u1 + (2*m2/(m1 + m2)) * u2
# Result: v1 = 0.0 m/s ✓Exam Tips
- Remember: Equal masses → velocity exchange in elastic collision
- Always verify with both conservation laws
- Fill-in answer: 0 (no decimal needed)
- Common trick: Distinguish elastic vs inelastic behavior
✅ Correct Answer: 0 m/s
First particle stops completely after elastic collision with equal mass at rest.
