physics- bullet?

Question

A rifle bullet with mass 8.00 g strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring. The impact compresses the spring 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cm

a) Find the magnitude of the block's velocity just after impact.
b) What was the initial speed of the bullet?

Answer 1

a) Kinetic energy of block = potential energy in spring

½ mv² = ½ kx²

Here m stands for combined mass (block + bullet), which is just 1 kg. Spring constant k is unknown, but you can find it from given data:

k = 0.75 N / 0.25 cm = 3 N/cm, or 300 N/m.

From the energy equation above, solve for v,

v = v √(k/m) = 0.15 √(300/1) = 2.598 m/s.

b) Momentum before impact = momentum after impact.
Since m = 1 kg, v = 2.598 m/s, p = 2.598 kg m/s. This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,

u = 2.598 kg m/s / 8 E-3 kg = 324.76 m/s.

Answer 2

The first problem can be solved with energy. Watch out, dont mix cgs with SI. I do problems in SI whenever possible. It is just what I'm used to. The newtons and kg are SI and the cm is cgs(yuck).
You get the potential energy of the spring by
E = 1/2 k d^2
In SI
E = joules
k = newtons per meter
d = meters

Ok now we can get the speed of the brick+bullet

E = 1/2 m v^2
E = joules
m = kg
v = meter per second

Now for the bullet speed we use conservation of momentum
momentum of bullet + momentum block =
momentum (block + bullet)
I think this is a one dimentional problem so we don't have to worry about the vector part.
It also looks like the block was not moving before the impact. So the block had zero momentum before impact.

So we can just say

speed (block + bullet) x mass (block + bullet) =
speed(bullet) x mass(bullet)

By now you have 3 of 4 of these terms so you can solve for the speed of the bullet.
God bless you in your efforts!

Answer 3

first use the information to find the factor k for the spring

F=kx

0.75=k .25 x 10^-2, k =300

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

v=x*sqrt(k/m)=.15sqrt(300/1)=2.598 m/s

use conservation to determine initial bullet velocity

m(total)v(initial)=m(bullet)v(bullet)+m(block)v(block)

1*2.6=.992*v(bullet); v(bullet)=2.619