Spring compression?

Question

All right, I don't even know where to start with this one. Do I use F = -kx? What formulas should I use after that, or at all? This is frustrating me.

"A 10.0 g bullet is fired horizontally into a 108 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 159 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 72.0 cm, what was the speed of the bullet at impact with the block?"

I don't think I'm supposed to use F = -kx to find the speed of impact, as F = ma and I don't know the acceleration. This, to me, doesn't sound right. What do I do?

Answer 1

First, use the law of conservation of momentum to find kinetic energy of the block and bullet
m*V = (m + M)*v
v = V*m/(m + M)
Now use the law of conservation of energy:
(m + M)v^2/2 = kx^2/2
x = v*sqrt((m + M)/k) = V*m/sqrt(k(m + M))
V = x*sqrt(k(m + M))/m = .72 * sqrt(159*0.118) / 0.010 = 1350 m/s

Thank you "prof" for your corrections.

Answer 2

[EDIT -- Alexy V was right; I was wrong, and so is madhukar. I assumed that all the kinetic energy was converted to potential energy; but that is not the case. Whenever two colliding bodies stick together after impact, it can be shown mathematically that it is an inelastic collision--that is, their kinetic energy is NOT conserved (i.e. some of it is converted to heat.). My equations did not take this into account.)

You _can_ use F=ma to solve the problem, but that would be the hard way.

The easier way is to use conservation of energy. For this, you need to know that the amount of potential energy in a compressed spring is:

PE = kx²/2

At the moment just prior to impact, there is no potential energy, and all the kinetic energy is in the bullet:

E_initial = (m_bullet)(v_bullet)²/2

At the moment of maximum compression, there is no kinetic energy ('cause everything's stopped momentarily); and the potential energy is all in the compressed spring:

E_final = kx² / 2

By conservation of energy:

E_initial = E_final
(m_bullet)(v_bullet)²/2 = kx² / 2

[EDIT -- The above is incorrect. The final energy consists of the PE of the spring plus the heat generated during the inelastic collision. Alexy V's analysis is correct.]

They give you "m_bullet" and "x". Just solve for "v_bullet".

(Notice that the mass of the block does not even matter).

[EDIT -- Actually it does. See Alexy V's analysis.]

Answer 3

Kinetic energy of the bullet, (1/2)mv^2, gets converted into the potential energy of the block/bullet/spring system due to compression of the spring which is (1/2)kx^2
=> (1/2)mv^2 = (1/2)kx^2
=> v
= x √(k/m)
= (0.72)*√(159/0.01) (all units in SI system)
= 90.8 m/s.
[Note: Mass of block is not needed here.]

Answer 4

I didn't rework Alexey's entire analysis, but his approach appears to be correct. However, he did commit at least one units error, which is why his bullet velocity came out unrealistically slow. His division by 10 grams at the very end should have been a division by 0.01 kilograms. You should check his work to make sure he didn't make similar errors elsewhere.

To avoid this kind of error, it helps me to remind myself that a Newton is defined as a kg•m/s².