Homework question on Simple Harmonic Motion (the bullet question)

In summary, the problem involves a 1.08 kg block attached to a horizontal spring with a spring constant of 3650 N/m. A 10.6 g bullet is fired into the block, which is initially at rest on a frictionless surface. The bullet sticks to the block and causes subsequent oscillations with an amplitude of 7.10 cm. The task is to find the initial speed of the bullet. Using conservation of momentum, the initial velocity of the block can be calculated. Then, using the equation for simple harmonic motion and the chain rule, the initial velocity of the bullet can be determined. Finally, setting this initial velocity equal to the previously calculated velocity of the block, the speed of the bullet can be solved
  • #1
hellokitty
19
0

Homework Statement



A 1.08 kg block is attached to a horizontal spring with spring constant 3650 N/m. The block is at rest on a frictionless surface. A 10.6 g bullet is fired into the block, in the face opposite the spring, and sticks.

What was the bullet's speed if the subsequent oscillations have an amplitude of 7.10 cm?

Homework Equations



This is the only formula I can think of but I do not have time?
Vx=-VtsinO=-Awsinwt

The Attempt at a Solution


I do not know which formula I could use to start the problem...

HELP!
 
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  • #2
Use conservation of momentum to find the initial velocity of the mass. Find w in terms of the mass and spring constant. If the motion is described by x(t)=A*sin(wt) then what is v(t)? Differentiate x(t). Then the velocity you found using conservation of momentum should equal v(0). Solve for A. Try it.
 
  • #3
hmm...i didnt get it right..

can some one show me how to do it step by step??
 
  • #4
Call t=0 the time the bullet hits the block. x(t)=A*sin(wt), v(t)=w*A*cos(wt). Put t=0 into that and conclude v(0)=w*A. Compute v(0) using conservation of momentum. Show us what you did.
 
  • #5
Okay so I solved for V(0)= w * A= 412.723

The conservation of momentum is: V0= ( m1+ m2/ m1) * Vf
Vf= 412.723/ 10.8
= 38.166

But that answer is wrong...so what do I do now?
 
  • #6
You really aren't showing enough of your work. I can't tell what you are actually doing. What did you get for w? From what you've got so far, it looks like Vf should be set equal to v(0)=w*A.
 
  • #7
I calculated the force required to push the block by the amplitude .071meters,
F=259.15N
then i calculated what the speed of the bullet would have to be to achieve this force, i got a strange answer of 24km/s which seems incredibly fast...
 
  • #8
That doesn't sound right. But then neither does you method of solving it.
 
  • #9
Hello,

I believe the initial Amplitude is one of hellokitty's givens,

"What was the bullet's speed if the subsequent oscillations have an amplitude of 7.10 cm?"

Why is A being solved for?

Edit: I just solved the problem. Perhaps just "solving" for amplitude left some confused, so I will leave some hints.
1. Final steps don't include use of momentum equations, but equations for simple harmonic motion. Think on the reasoning behind this hint, is the ending momentum "constant" or in other words, is there a true "final" portion of there scenario, such as there is when two billiard balls hit each other?
2. When differentiating the position equation for SHM, don't forget the chain rule! (There is an important constant inside the trig. function of x(t))
3. After solving for the initial velocity of the mass and plugging that into v(t), look at the variable you wish to solve for and ask yourself how you may get rid of any other unknown constant.

It may be helpful to think of this as a three-stage momentum problem involving SHM.
You have:
Stage 1: The bullet is moving, block is at rest.
Stage 2. The instant when the block is hit by the bullet
Stage 3: System is now modeled by SHM; momentum is constantly changing with respect to the block's distance from the equilibrium point.

I hope this was helpful :)
 
Last edited:

1. What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion in which a system or object oscillates back and forth around a central equilibrium point. It is characterized by a sinusoidal or cosine wave pattern, with the object moving at a constant frequency and amplitude.

2. How is Simple Harmonic Motion related to the bullet question?

The bullet question typically refers to a scenario where a bullet is fired from a gun and travels through the air, eventually coming to a stop. This motion can be described as a type of simple harmonic motion, as the bullet experiences a restoring force (air resistance) that causes it to oscillate back and forth until it comes to a rest.

3. What factors affect the period of Simple Harmonic Motion?

The period of SHM is affected by three main factors: the mass of the object, the stiffness of the system, and the amplitude of the motion. A heavier object will have a longer period, while a stiffer system or larger amplitude will result in a shorter period.

4. How can we calculate the velocity of the bullet in Simple Harmonic Motion?

The velocity of the bullet in SHM can be calculated using the equation v = ωA cos(ωt), where ω is the angular frequency of the motion, A is the amplitude, and t is the time. This equation describes the velocity of a particle in SHM as a function of time.

5. What real-life examples exhibit Simple Harmonic Motion?

SHM can be observed in many real-life examples, such as the motion of a pendulum, a mass on a spring, or the motion of a tuning fork. It is also present in more complex systems, such as the motion of planets around the sun or the oscillation of sound waves.

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