Linear Collisions; Conservation Law

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SUMMARY

The discussion focuses on a collision experiment involving two masses, A (100 g) and B (200 g), on a horizontal air track with springs. The speed of mass A as it leaves spring 1 is calculated to be 0.71 m/s. After colliding with mass B, the velocity of mass B is determined to be 0.25 m/s to the right. The maximum compression of spring 2 during the collision is found to be 2.5 cm. The conservation of energy and momentum principles are applied to solve the problems, highlighting the importance of accurate unit conversions.

PREREQUISITES
  • Understanding of conservation of energy (Ep = Ek)
  • Knowledge of conservation of momentum (p = mv)
  • Familiarity with spring constant calculations (k = 20 N/m)
  • Ability to perform unit conversions (e.g., grams to kilograms, centimeters to meters)
NEXT STEPS
  • Study the principles of conservation of momentum in elastic collisions
  • Learn about energy conservation in mechanical systems
  • Explore the effects of spring constants on motion dynamics
  • Practice unit conversion techniques in physics problems
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Students studying physics, particularly those focusing on mechanics and collision theory, as well as educators looking for practical examples of conservation laws in action.

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Homework Statement


The student sets up a collision experiment with two masses on a horizontal air track, with two springs on each end. The mass of A is 100 g and B is 200 g. Each spring has a constant of 20 N/m. Spring 1 (left side) was initially compressed 5 cm
to start mass A moving.

a. Determine the speed of mass A as it leaves spring 1 [0.71 m/s]
b. If A has a velocity of 20 cm/s to the right after it collides with B, determine the velocity of mass B. [0.25 m/s right]
c. Determine the maximum compression of spring 2 when mass B collides with it. [2.5cm]

Homework Equations


Ep = 1/2kx2
p = mv

The Attempt at a Solution


I solved part a readily using the conservation of energy law (Ep = Ek). When I tried to solve part b using conservation of momentum, however, I obtained an incorrect answer:

Total momentum before = Total momentum after
m1v1+m2v2=m1v3+m2v4
(m1v1+m2v2-m1v3)/m2=v4
[(0.1kg)(0.71m/s)+(0.2kg)(0)-(0.1kg)(0.02m/s)]/(0.2kg)= v4
v4 = 0.35 m/s right

Am I reading the question wrong? Are my formulae incorrect?​
 
Last edited:
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20 cm/s would be 0.20 m/s, not 0.02 m/s
 
Hahahaha thanks for your help; I'll have to be more diligent with unit conversions next time!
 

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