How many meters does the spring compress?

[linnus]

Homework Statement

A 10.5 kg block, attached to the left end of a horizontal massless spring, sits on a frictionless table. The right end of the spring is attached to a vertical piece of wood that is firmly nailed to the table. A 0.0500 kg projectile is fired, from left to right, into the block at 85.5 m/s and stops inside it (this is a completely inelastic collision). The spring constant is k = 105 N/m. How many meters does the spring compress? The potential energy due to the compression of the spring can be calculated with the following formula: PE = (1/2)kx^2.

Homework Equations

1/2mv^2
1/2kx^2
momentum before=momentum after

The Attempt at a Solution

uh...
Why can't you do...
1/2mv^2=1/2kx^2?
x=1.866...which isn't right

 

[dwintz02]

Break down the problem into 2 steps. The bullet is going to become embedded in the block. For inelastic collisions, energy conservation is generally not useful because some of the kinetic energy of the bullet is dissipated by friction or turned into heat, and neither of these are easy to calculate. Therefore, use conservation of momentum to calculate the velocity of the block with the bullet embedded in it IMMEDIATELY after the bullet is embedded (ignore the spring for now). Then, when you have that velocity, you can calculate the kinetic energy of the block/bullet (sum the masses also, do you see why?) combo and then use energy conservation to solve for the compression of the spring.

 

[ogb p]

.

I understand your confusion with the given information. However, it is important to consider the conservation of momentum in this situation. The projectile has momentum before it collides with the block, and after the collision, the combined mass of the block and projectile will have momentum. This means that the momentum before must equal the momentum after. We can use this information to find the velocity of the combined mass after the collision, using the equation:

mv projectile = (m block + m projectile) v combined

Solving for v combined, we get:

v combined = (m projectile / (m block + m projectile)) v projectile

Plugging in the given values, we get:

v combined = (0.0500 kg / (10.5 kg + 0.0500 kg)) (85.5 m/s) = 0.409 m/s

Now, we can use this velocity to calculate the compression of the spring using the equation:

1/2mv^2 = 1/2kx^2

Solving for x, we get:

x = √(mv^2 / k)

Plugging in the values, we get:

x = √((0.0500 kg)(0.409 m/s)^2 / 105 N/m) = 0.0055 m = 5.5 mm

Therefore, the spring compresses by 5.5 mm in this scenario. It is important to consider all variables and equations in a situation like this to arrive at the correct answer.