Question:
\(n\) small balls, each of mass \(m\) impinge elastically each second on a surface with a velocity \(u\), then the force experienced by the surface in one second, will be
\(n\) small balls, each of mass \(m\) impinge elastically each second on a surface with a velocity \(u\), then the force experienced by the surface in one second, will be
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For inelastic collision, change in momentum = \(mu\) (ball stops). For elastic, it's \(2mu\).
Updated On: Apr 23, 2026
- \(4\ \text{mn}u\)
- \(2\ \text{mn}u\)
- \(1.5\ \text{mn}u\)
- \(0.8\ \text{mn}u\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Force = rate of change of momentum. For elastic collision with stationary surface, velocity reverses.
Step 2: Detailed Explanation:
Initial momentum of one ball = \(mu\) (towards surface).
Final momentum after elastic collision = \(-mu\) (away from surface).
Change in momentum for one ball = \(mu - (-mu) = 2mu\).
For \(n\) balls per second, total change in momentum per second = \(n \times 2mu = 2mnu\).
Force = rate of change of momentum = \(2mnu\).
Step 3: Final Answer:
Thus, force = \(2\ \text{mn}u\).
Force = rate of change of momentum. For elastic collision with stationary surface, velocity reverses.
Step 2: Detailed Explanation:
Initial momentum of one ball = \(mu\) (towards surface).
Final momentum after elastic collision = \(-mu\) (away from surface).
Change in momentum for one ball = \(mu - (-mu) = 2mu\).
For \(n\) balls per second, total change in momentum per second = \(n \times 2mu = 2mnu\).
Force = rate of change of momentum = \(2mnu\).
Step 3: Final Answer:
Thus, force = \(2\ \text{mn}u\).
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