Conservation of Momentum and gravity

In summary: So the initial speed is just v. In summary, a student performed a ballistic pendulum experiment and obtained the following data: the maximum height reached by the pendulum was 3 cm, at this height the pendulum made an angle of 36.9◦, the mass of the bullet used was 97 g, the mass of the pendulum bob was 788 g, and the acceleration of gravity was 9.8 m/s2. The student attempted to solve for the final speed of the system using the equations (1/2)(m1+m2)v^2 (final) = (m1+m2)gh and (m1+m2)V (initial) = (m1+m2)v(final)
  • #1
jcjp
1
0

Homework Statement


A student performs a ballistic pendulum experiment using an apparatus similar to that shown in the figure. Initially the bullet is fired at the block while the block is at rest (at its lowest swing point). After the bullet hits the block, the block rises to its highest position, see dashed block in the figure, and continues swinging back and forth. The following data is obtained: the maximum height the pendulum rises is 3 cm, at the maximum height the pendulum subtends an angle of 36.9◦, the mass of the bullet is 97 g, and the mass of the pendulum bob is 788 g. The acceleration of gravity is 9.8 m/s2 .


Homework Equations



I thought this was the equation I needed to use: (1/2)(m1+m2)v^2 (final) =( m1 +m2)gh
(to solve for v(final))

Along with: (m1+m2)V (initial)= (m1 +m2) v(final)

The Attempt at a Solution



I continually get around 2.277, yet I know that's wrong.

Also, I don't really know where the trig is applied...
 
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  • #2
jcjp said:

Homework Statement


A student performs a ballistic pendulum experiment using an apparatus similar to that shown in the figure. Initially the bullet is fired at the block while the block is at rest (at its lowest swing point). After the bullet hits the block, the block rises to its highest position, see dashed block in the figure, and continues swinging back and forth. The following data is obtained: the maximum height the pendulum rises is 3 cm, at the maximum height the pendulum subtends an angle of 36.9◦, the mass of the bullet is 97 g, and the mass of the pendulum bob is 788 g. The acceleration of gravity is 9.8 m/s2 .

The Attempt at a Solution



I continually get around 2.277, yet I know that's wrong.

Also, I don't really know where the trig is applied...

The question doesn't say what you have to compute and I only know that you "get around 2.277 and that it's wrong" (what quantity? units?) so I'll just outline the general approach.

m - mass of the bullet
M - mass of the block
v - speed of the bullet
V - speed of the block+bullet system
h - distance the system rises
K - initial kinetic energy of the system after impact
U - potential energy of the system at the highest position

To get the speed of the system after the impact just apply the conservation of momentum:

mv = (m + M)V

Then conservation of energy:

K = U

K = 1/2 * (m + M)V^2

U = (m + M)gh

The rest is just rearranging and substituting.
 
  • #3
jcjp said:
...

Along with: (m1+m2)V (initial)= (m1 +m2) v(final)

...

This looks to be the problem, since only the bullet is moving initially, the block is at rest.
 

FAQ: Conservation of Momentum and gravity

What is the Conservation of Momentum?

The Conservation of Momentum is a fundamental law in physics that states that the total momentum of a closed system remains constant. This means that in the absence of external forces, the total momentum before an event will be equal to the total momentum after the event.

How does Gravity affect the Conservation of Momentum?

Gravity is a force that acts on all objects with mass and is one of the four fundamental forces in the universe. It plays a crucial role in the Conservation of Momentum by exerting a force on objects and causing them to accelerate, which in turn affects their momentum. The gravitational force can either increase or decrease an object's momentum depending on its direction and strength.

Can the Conservation of Momentum be violated?

No, the Conservation of Momentum is a fundamental law of physics and has been extensively tested and proven to hold true in all situations. While it may seem like momentum can be violated in certain scenarios, it is actually just being transferred and conserved in different forms, such as kinetic energy or potential energy.

How does the Conservation of Momentum apply to collisions?

In collisions, the Conservation of Momentum states that the total momentum before the collision will be equal to the total momentum after the collision. This means that the sum of all the momenta of the objects involved will remain constant. This law can be used to analyze and predict the outcomes of different types of collisions, such as elastic and inelastic collisions.

What are some real-world applications of the Conservation of Momentum and gravity?

The Conservation of Momentum and gravity play a crucial role in many real-world applications, such as rocket launches, satellite orbits, and sports. In rocket launches, the momentum of the expelled gases propels the rocket in the opposite direction, following the Conservation of Momentum. In sports, understanding the conservation of momentum can help athletes improve their performance, such as in long jump or javelin throw events.

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