How to find amplitude in SHM problem

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SUMMARY

The discussion centers on calculating the amplitude of a mass-spring system following a collision. A 120g mass rolls down a frictionless hill, achieving a speed of 4.2 m/s before colliding with a 3.00g mass attached to a spring with a constant of 30 N/m. The final velocity after the collision is determined to be 1.2 m/s, and the angular frequency (w) is calculated as 8.45 rad/s. The correct amplitude (A) is confirmed to be 0.266 m, leading to the motion equation x(t) = 0.266sin(8.44t).

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  • Understanding of conservation of momentum in collisions
  • Familiarity with simple harmonic motion (SHM) equations
  • Knowledge of angular frequency calculations (w = k/m^1/2)
  • Ability to manipulate trigonometric functions in motion equations
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  • Study the principles of conservation of momentum in elastic and inelastic collisions
  • Learn about the derivation and application of simple harmonic motion equations
  • Explore the effects of mass and spring constant on oscillation frequency
  • Investigate energy transformations in mass-spring systems
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Homework Statement


A mass of 120g rolls down a frictionless hill reaching a speed of 4.2 m/s and collides with another mass 3.00g attached to a spring of constant 30 N/m. The two masses stick together and enter into periodic motion. What is the equation for the motion?

Homework Equations


m1v1+m2v2=mtV
x(t)=Acos(wt+phi)
w=k/m^1/2
Vmax=wA

The Attempt at a Solution


I found Vfinal=1.2 m/s
Also found w=8.45 rad by w=k/m^1/2
I thought A would be 0.142 but in the answer A=0.266
the answer is: x(t)=0.266sin(8.44t)
 
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I guess you meant to type 300 g rather than 3.00 g for the second mass.

Your work looks good to me. I agree that A = 0.142 m.

[EDIT: I'm assuming that the first mass reaches the bottom of the hill and then moves horizontally before hitting the spring as shown below.]

(If there is no friction on the hill, the 120 g mass would slide rather than roll down the hill.)
 

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