Calculate Initial Bullet Speed in SHM with Block-Spring System

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SUMMARY

The discussion centers on calculating the initial speed of a bullet in a block-spring system undergoing simple harmonic motion (SHM). The bullet, weighing 5.4 g, is fired into a 0.50 kg block attached to a spring with a frequency of 9.5 Hz and an amplitude of 15 cm. The initial calculations incorrectly used the displacement equation instead of the correct differential equation for SHM. The correct initial speed of the bullet is determined to be 14.6 m/s after applying the appropriate equations.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with conservation of momentum principles
  • Knowledge of angular frequency calculations
  • Ability to differentiate displacement functions in physics
NEXT STEPS
  • Study the principles of conservation of momentum in inelastic collisions
  • Learn about the equations governing simple harmonic motion (SHM)
  • Explore the relationship between frequency, amplitude, and angular frequency in SHM
  • Practice solving problems involving differential equations in physics
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Students studying physics, particularly those focusing on mechanics and simple harmonic motion, as well as educators seeking to clarify concepts related to momentum and SHM calculations.

lando45
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My teacher set us a bunch of pendulum/SHM questions, and I managed to crack them all except for this one which has been bothering me:

"A 5.4 g bullet is fired horizontally into a 0.50 kg block of wood resting on a frictionless table. The block, which is attached to a horizontal spring, retains the bullet and moves forward, compressing the spring. The block-spring system goes into SHM with a frequency of 9.5 Hz and an amplitude of 15 cm. Determine the initial speed of the bullet."

I found the following diagram to help me calculate my answer but I came up with the wrong answer.

shm.gif


Here's what I did:

Displacement (y) = Amplitude (A) x sin (Angular Frequency [w] x Time [t])
y = 0.15 x sin ([2 x Π x f] x [1 / f])
y = 0.15 x sin (59.7 x 0.105)
y = 0.15 x 0.109
y = 0.0164
So in time 0.105 seconds the block w/ spring moved 0.0164 metres which is equivalent to 0.156ms^-1.

I then plugged this value into the motion equation m1v2 = m2v2 and got the following:

0.0054u = 0.5054v
v = 0.156
0.0054u = 0.5054 x 0.156
0.0054u = 0.079
u = 14.6ms^-1

I tried submitting this answer but it was wrong...can someone please help me out!
 
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EDIT: OK I just solved it, I realized I was using the wrong equation, I should have been using the differential of the y displacement function. Thanks anyway.
 
Sometimes the universe do listen to our problems and have pity on us.
 

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