# Stumped with Physics Questions

1. Oct 3, 2005

### Ryan231

I'm really stumped on 3 of my physics questions.. Ive been able to find a few things but theres so many equations and variables I've gotten myself lost.. if anyone could help it would be greatly appreciated!

A 25.0 g bullet strikes a 0.600 kg block attached to a fixed horizontal spring whose spring constant is 6700 N/m and set it into vibration with an amplitude of 21.5 cm. What was the speed of the bullet before impact if the two objects move together after the impact?

-not sure what I need to do in this case? tried finding the total energy when v=0 and x= +- A, but that lead me no where

The other question...
The length of a simple pendulum is 0.66 m, the pendulum bob has a mass of 310 grams, and it is released at an angle of 12º to the vertical. (a) With what frequency does it vibrate? Assume SHM. (b) What is the pendulum bob’s speed when it passes through the lowest point of the swing? (c) What is the total energy stored in this oscillation, assuming no losses?

Found F= 0.9Hz not sure how to incorporate the angle into the equation to solve the rest?

Finally...
If two successive overtones of a vibrating string are 280 Hz and 350 Hz, what is the frequency of the fundamental?

THANKS to anyone who can help me out!

2. Oct 3, 2005

### Ryan231

One the first question I tried ma = -kx where k is the amplitude.. but it gave me an answer of a = -5,762,000 Really stumped here..

3. Oct 3, 2005

### ZapperZ

Staff Emeritus

First of all, we can't be sure if we can use conservation of energy of the bullet as the situation before, and the bullet+block right after impact. This is because we don't know if this is an elastic collision or not. Typically, when two objects stick to each other, it is not an elastic collision, and thus KE isn't conserved. But momentum does! So use that.

You have momentum of the bullet as m1v1.

At the instant it hits and gets imbedded into the block, you have (m1+m2)v2. This is the momentum after the collision.

Now, at this point, you CAN use conservation of energy, because in an oscillation, energy is conserved at every point of the oscillation. So you have the initial KE of the bullet+block as 1/2 (m1+m2)v2^2 (I'm too lazy to format this in LaTex right now so I hope you can follow this). When it is completely compressed by the amount given (i.e. the amplitude A), all the KE has been converted into the spring PE, i.e. 1/2 kA^2.

So now what do we have?

m1v1 = (m1+m2)v2
1/2 (m1+m2)v2^2 = 1/2 kA^2

You have two unknowns (v1 and v2), and two equations. You can solve this now, and the question asked for v1.

Zz.

4. Oct 3, 2005

### Ryan231

I'm pretty bad at physics overall so I'm not sure what you mean at this point:
"You have momentum of the bullet as m1v1."

I would assume you mean the momentum is 25g*velocity1

thus making it...
m1v1 = (.025kg+.6kg)v2
.5*(.025+.6)v2^2 = .5kA^2

.5*6700*(.215m^2) = 154.85

.5*(.625)v2^2 = 154.85
v2 = 25.26

thus... m1v1 = (.625kg)(25.26)
speed of bullet = 15.7875m/s?

I hope this is somewhat close??

5. Oct 3, 2005

### ZapperZ

Staff Emeritus
Hint: when solving ANY physics problems, DO NOT substitute in the numbers way in the beginning. If you do that, you will be stuck very quickly. Assign SYMBOLS to all the variables and quantities. You'll see that in many cases, some of these will cancel out. If you're in the habit of always plugging in numbers at the very beginning, you'll realize you don't have the values for several quantities without realizing that they'll cancel. This is no good.

Solve the problem algebraically first (that's why we learn algebra). Only plug in numbers at the end.

Zz.

6. Oct 3, 2005

### Ryan231

Ok I'm not great at doing it the other way but I'll give it a shot so...

m1v1 = (m1+m2)v2
1/2 ( m1 + m2)v2^2 = 1/2kA^2

Sub m1v1 = (m1+m2)v2 into 1/2 (m1+m2)v2^2 = 1/2kA^2
= 1/2 m1v1^2 = 1/2kA^2
Cancel out ^2's and 1/2's
m1v1 = kA
.025kg v1 = 6700(.215m)
v1 = 57,620???

This is an answer I got before but seems very wrong.. I've been trying this question for 3 hours and keep going in circles...

7. Oct 3, 2005

### Ryan231

i guess the amplitude of vibration is the full length of compression to the spring, in which case there will be potential energy of .5kx^2 stored in the spring. that's equal to the initial kinetic energy of the bullet since there's no initial kinetic energy in the block.

so .5kx^2=mv^2, where m = mass of bullet and v = initial speed of bullet

.5*6700*0.215^2=0.025*v^2

v=78.7 m/s

can someone with experience that knows this stuff just tell me if I'm at least right? then I can work backwards and actually piece this together