Calculating Bullet Speed Before Impact for Vibration Problem

[confusedaboutphysics]

A 20.0 g bullet strikes a 0.600 kg block attached to a fixed horizontal spring whose spring constant is 6.74 103 N/m and sets it into vibration with an amplitude of 21.2 cm. What was the speed of the bullet before impact if the two objects move together after impact?

I've been stuck on this question for days! help please!

 

[Päällikkö]

For the block+bullet -system to oscillate with the given amplitude, what must the (initial kinetic) energy of the system be?

During impact energy is not conserved, though.

 

[quicknote]

To calculate the bullet speed before impact, we can use the conservation of momentum principle. This states that the total momentum of a system before and after a collision remains constant. In this case, the system consists of the bullet and the block attached to the spring.

We can set up the equation as follows:

m1v1 + m2v2 = (m1 + m2)v

Where m1 is the mass of the bullet, v1 is the speed of the bullet before impact, m2 is the mass of the block, v2 is the speed of the block before impact, and v is the speed of the two objects moving together after impact.

We know the mass of the bullet (m1 = 20.0 g = 0.020 kg) and the mass of the block (m2 = 0.600 kg). We also know the spring constant (k = 6.74 x 10^3 N/m) and the amplitude of vibration (A = 21.2 cm = 0.212 m).

To solve for v1, we need to first find v2. We can use the equation for the period of a simple harmonic motion to find v2:

T = 2π√(m/k)

Where T is the period of vibration (T = 2π/ω, where ω is the angular frequency), m is the mass of the block, and k is the spring constant.

Substituting the values, we get:

T = 2π√(0.600 kg / 6.74 x 10^3 N/m) = 0.018 s

Next, we can use the equation for the speed of an object in simple harmonic motion to find v2:

v2 = ωA

Where ω is the angular frequency (ω = 2π/T) and A is the amplitude of vibration. Substituting the values, we get:

v2 = (2π/0.018 s)(0.212 m) = 74.2 m/s

Now that we have v2, we can plug it into the conservation of momentum equation to solve for v1:

(0.020 kg)v1 + (0.600 kg)(74.2 m/s) = (0.020 kg + 0.600 kg)v

Simplifying, we get:

v1 = (0.600 kg)(74.2