Springs, Simple Harmonic Motion

[ceeforcynthia]

1. A 25.0g bullet strikes a .600kg block attached to a fixed horizontal spring whose spring constant is 6.70x10^3 and sets it into vibration with an amplitude of 21.5cm. What is the speed of the bullet before impact if the two objects move together after impact?2. V=A$$\sqrt{k/m}$$
3. I used the equation above, adding the masses together (.625kg). I got the velocity to be 22.26m/s... but now i can't figure out how to get the speed of the bullet.


The answer in the book says that the speed of the bullet is 557m/s.

 

[Astronuc]

One determines the maximum spring deflection from the amplitude, and from that the stored energy in the spring at max deflection, which by conservation of energy (and neglecting friction) should equal the initial kinetic energy of the bullet and block.

From the KE of the bullet and block, one can determine the velocity of the combination.

Since the bullet striking the block is an example of an inelastic collision, what about the momentum of the bullet and block?

 

[ogb p]

There are a few steps missing in the provided solution, which is why the calculated speed of the bullet does not match the answer in the book. Let's break down the problem and see how we can arrive at the correct answer.

First, we need to understand the concept of simple harmonic motion. When a block attached to a spring is displaced from its equilibrium position, it experiences a restoring force that is proportional to its displacement. This results in the block oscillating back and forth around its equilibrium position. The equation V=A\sqrt{k/m} is the formula for finding the velocity (V) of the block in simple harmonic motion, where A is the amplitude of the motion, k is the spring constant, and m is the mass of the block.

In this problem, the block is initially at rest, and the bullet strikes it, causing it to move with a certain velocity. After the impact, the two objects move together as one unit. This means that the total mass of the system is now 0.625kg (0.600kg block + 0.025kg bullet).

To find the velocity of the bullet before impact, we can use the principle of conservation of momentum. This states that the total momentum of a system before and after a collision remains constant. Mathematically, we can express this as m1v1 = m2v2, where m1 and v1 are the mass and velocity of the bullet before impact and m2 and v2 are the mass and velocity of the combined system after impact.

We can rearrange this equation to solve for the velocity of the bullet before impact, which gives us v1 = m2v2/m1. Plugging in the values, we get v1 = (0.625kg)(0.22.26m/s)/0.025kg = 557m/s, which matches the answer in the book.

In conclusion, the speed of the bullet before impact can be calculated by using the principle of conservation of momentum and the formula for simple harmonic motion. It is important to carefully consider all the given information and apply the appropriate equations to arrive at the correct solution.