Please help with physics problem!!!?

Question

A bullet with mass 4.66 g is fired horizontally into a 1.929 kg block attached to a horizontal spring. The spring has a constant 6.34 102 N/m and reaches a maximum compression of 6.38 cm.
(a) Find the initial speed of the bullet-block system.
vi = m/s

(b) Find the speed of the bullet.
vbullet = m/s

I would really appreciate it if someone could please help me with this problem. I have been stuck on it for a while and I have asked several classmates as well, which are also stuck. Thank you so much in advance!

This is the exact problem that I copied and pasted.

Answer 1

I think you're spring constant is 6.34 * 10² N/m, so that's what I'm going to use.

The is definitely an energy problem. We are going to ignore friction. Since we're ignoring friction, no energy is lost.

Ei = Ef

When the block starts out, it has no PEi because the spring is not compressed or stretched. When the block finally reaches its maximum compression on the spring, it's not moving, so KEf = 0.

Therefore,

KEi = PEf

½m v² = ½k x²

Plug and chug :).. The ½ cancels out, leaving

v = sqrt(k x² / m)

k = spring constant
x = distance spring was compressed
m = mass of bullet + block

v = sqrt(6.34 * 10² * (.0638)² / (.00466 + 1.929))
v = 1.1553 m/s

That's the speed of the bullet/block system.


Now for the speed of the bullet.. This is definitely a momentum conservation problem. Momentum is ALWAYS conserved!!!

Pi = Pf

Initially, the block is not moving, so it has no momentum.

m vb = (m + M) v

vb = (m + M)v / m

vb = (.00466 + 1.929)*1.1553 / .00466

vb = 479.368 m/s


Make sure you convert everything to standard units before doing any calculations!

Answer 2

The given data sounds suspicious. The spring constant of "6.34 102" N/m looks peculiar. It has an unnatural space in between the 4 and the 1, and it seems as if it has too many decimal digits. Can you recheck the given conditions and restate what the spring constant is?

Answer 3

Not going to do the math, but this is how you would do it. You are going to use two concepts. Part (a) is conservation of energy. Part (b) is conservation of momentum.

(a) conserve the kinetic energy of block+embedded bullet with potential energy in the spring:

You know the kinetic energy of the block+embedded bullet at the equals the total stored energy in the spring at maximum compression (assuming no friction).

KE of block+bullet =0.5*(m_bullet+m_block)*V_block^2
PE of compressed spring=0.5*K_spring*delta_X^2

where delta_X=distance that sprince is compressed.

Set the two equal to each other and solve for the velocity of the block with the embedded bullet. You know everything else.

(b) conservation of momentum

The momentum of the block with the embedded bullet right after it hit has to equal the momentum of the bullet going into the block.

P_bullet+block = (m_block+m_bullet) * V_block
P_bullet = m_bullet * V_bullet

Set the two to each other and solve for V_bullet. You found V_block in part (a).