Phys4AS17cceron: April 26, 2017 Lab: Ballistic Pendulum

Lab: Ballistic Pendulum

Chris Ceron, Amy, John, Ricardo

Goal of Lab

To use conservation of momentum and energy to determine the firing speed of a ball for a spring-loaded gun.

Apparatus used in lab

Theory/Introduction

Within our apparatus there are in theory no external forces acting on our system. This allows for us to calculate the initial velocity using the mass and final velocity of the system. Our equations for momentum are as follows:

Initial Momentum:

$p_{i} = m_{ball}v_{ball_{i}}+m_{block}v_{block_{i}}$

Final Momentum:

$p_{f} = (m_{ball}+m_{block})v_{f}$

Since momentum is conserved, initial and final momentum are equal:

$m_{ball}v_{ball_{i}}+m_{block}v_{block_{i}} = (m_{ball}+m_{block})v_{f}$

Using this equation, we could solve for the final velocity of the ball to later use in our conservation of energy equation:

$v_{f}=\frac{m_{ball}v_{ball_{i}}+m_{block}v_{block_{i}}}{(m_{ball}+m_{block})}$

Since we assume that energy is also conserved, we can derive a formula for the energy in our system:

$KE_{i} + U_{i} = KE_{f}+U_{f}$

where KE is the Kinetic Energy and U is our potential energy.

Before the collision, the energy within our system is all in the form of Kinetic Energy. After the collision, that energy is transferred to potential energy:

$\frac{1}{2}(m_{ball}+m_{block}){v_{f}}^{2} + 0 = 0 + (m_{ball}+m_{block})gh$

The height used in our potential energy can be found by taking the length (L) of the string, and subtracting it by the x component (Lcos $\theta$ ) of the string after the collision.

If we use the final velocity we found through momentum and simply, the result is:

$v_{ball_{i}} = \frac{m_{ball}}{m_{ball}+m_{block}}\sqrt{2g(L-Lcos\theta)}$

To verify our initial velocity, we move the nylon block out of the ball's trajectory and fire the ball at 0 degrees. We can place a piece of carbon paper where we expect the ball to land based on the velocity value found in part one of the lab. This becomes a projectile motion problem. We can use the height at which the ball is fired and the distance the ball travels in the x direction to calculate the initial velocity.

$\Delta y = v_{y_{i}}t+\frac{1}{2}a_{y}t^2 \: \: ,\: \: \Delta x = v_{x_{i}}t+\frac{1}{2}a_{x}t^2$

We use the first equation to solve for t. Since the ball was fired horizontally, the initial velocity in the y-direction is 0. We can substitute t in the second equation and solve for the initial velocity. Since there is no force acting on the ball horizontally, we know that the acceleration in the second equation is equal to 0.

$t=\sqrt{\frac{2\Delta y}{g}} \Rightarrow \Delta x = v_{x_{i}}\sqrt{\frac{2\Delta y}{g}} \Rightarrow v_{x_{i}} = \frac{\Delta x}{\sqrt{\frac{2\Delta y}{g}}}$

To calculate our uncertainty, we look at each calculation separately. For the collision part of our lab, we take the natural log of our equation:

$\ln V_{ball_{i}} = \ln\left(\frac{m_{ball}}{m_{total}}\sqrt{2g(L(1-cos\theta))}\right)$

We can use a property of natural log to expand everything:

$\ln v_{ball_{i}} = \ln m_{ball}- \ln m_{total} + \frac{1}{2}\left(\ln 2 + \ln g + \ln L + \ln (1-cos\theta)\right)$

When we take the derivative, the constants will disappear. Our constants in this case are ln(2) and ln(g)

$\frac{dv_{ball_{i}}}{v_{ball_{i}}} = \frac{dm_{ball}}{m_{ball}}-\frac{dm_{total}}{m_{total}}+\frac{dL}{2L}+\frac{d(1-cos\theta)}{2(1-cos\theta)}$

We can then use the propagated uncertainty equation to find the total uncertainty:

$\frac{dv_{ball_{i}}}{v_{ball_{i}}} = \sqrt{\left(\frac{dm_{ball}}{m_{ball}}\right)^{2}+\left(\frac{dm_{total}}{m_{total}} \right )^{2}+\left (\frac{dL}{2L}\right)^{2} +\left(\frac{d(1-cos\theta)}{2(1-cos\theta)}\right)^{2}}$

To find the propagated uncertainty in our verification portion of the lab, we do the same thing.

$\ln v_{ball_{i}} = \ln \Delta x -\frac{1}{2}\left(\ln 2 + \ln \Delta y - \ln g \right )$

$\frac{dv_{ball_{i}}}{v_{ball_{i}}} = \frac{d \Delta x}{\Delta x} - \frac{d \Delta y}{2\Delta y}$

$\frac{dv_{ball_{i}}}{v_{ball_{i}}} = \sqrt{\left(\frac{d \Delta x}{\Delta x} \right )^{2} + \left(\frac{d \Delta y}{2\Delta y} \right )^{2}}$

Procedure
We begin our lab by recording the mass of the ball we use to fire into the nylon block. We then measure the length of the string that is attached to the nylon block. We ensure that our ballistic pendulum is leveled so that the ball is fired straight into the block. This is important because the ball could lose momentum if the ball hits the sides of the block before being fully embedded. Our apparatus allows for different firing rates, so we chose to use three notches. After the collision, both the ball and block act as one and travel in the same velocity. Due to the collision, the block move in the direction of the ball and goes up a certain height. We can find the height by subtracting the length of the string and the x component of the string as mentioned in the theory/introduction. We repeated this part of the experiment five times and used the average height.

For the verification portion of the lab, we move the nylon block out of the ball's trajectory as mentioned in the theory/introduction. We fired the ball three times using three notches, and used the average distance from the recordings made by the carbon paper.

Data/Calculations

mass of the ball: 0.0076 kg +/- 0.0001 kg
mass of the block: 0.0795 +/- 0.0001 kg
length of the string (L): 0.195 m +/- 0.001 m

Trial #	Angle (degrees)
1	35.5
2	35.5
3	34
4	34.5
5	36.5
	Average: 35.2

Carbon marking from the verification portion of lab

Distance in x direction that ball traveled

Data for the verification portion of lab:

height from ground to the point of fire: 0.945 m +/- 0.001 m

length from ground below point of fire to the center of impact: 3.34 m +/- 0.001 m

Phys4AS17cceron

Tuesday, May 9, 2017

April 26, 2017 Lab: Ballistic Pendulum

No comments:

Post a Comment