Simple harmonic motion and amplitude

AI Thread Summary
A 2.10-kg block suspended from a spring with a spring constant of 280.0 N/m is struck by a 0.070-kg bullet traveling at 220.0 m/s, embedding itself in the block. To find the amplitude of the resulting motion, the conservation of momentum should be applied to determine the initial velocity of the combined block and bullet immediately after the collision. The kinetic energy of the bullet before the collision can be compared to the mechanical energy of the block-spring-bullet system after the collision. The discussion emphasizes using the equations for kinetic and potential energy to solve for amplitude and the fraction of kinetic energy converted to mechanical energy. Understanding these principles is crucial for solving the problem effectively.
Dotty21690
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Homework Statement



A 2.10-kg block is suspended from a spring with a spring constant of 280.0 N/m. A 0.070-kg bullet is fired into the block from directly below with a speed of 220.0 m/s and is embedded in the block. Find the amplitude of the subsequent motion.

-What fraction of the original kinetic energy of the bullet appears as mechanical energy in the system of block-spring-bullet?

Homework Equations



we are learning about the oscillations and energy. For this question I'm thinking I would need to use E= Kinetic energy + potential energy... (1/2)KA2=(1/2)mv2 + (1/2)Kx2
but I don't know what to do to get started, I am soo lost!

The Attempt at a Solution

 
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Start by finding the speed of the "bullet + block" immediately after the bullet embeds itself in the block. What's conserved during that 'collision'?
 
I suggest using conservation of momentum on the collision between the bullet and block to find the initial velocity upwards of the block.
 
thanks! but then how would I find part b? would I find the kinetic energy of the bullet and the mechanical energy of the system(in which I would plug back in my value for the amplitude?)
 
Dotty21690 said:
thanks! but then how would I find part b? would I find the kinetic energy of the bullet and the mechanical energy of the system(in which I would plug back in my value for the amplitude?)
Compare the KE before and after the collision.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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