Ballistic bullet pendulum conservation of momentum problem

In summary, a 15 g bullet with a speed of 425 m/s is fired into a 2.55 kg block hanging by a 2.0 m long string. To calculate the angle when the system reaches its maximum height, the equations m1v1=(m1+m2)vf and (1/2)mv^2=mgh are used. After solving for the final velocity and height, the equation L-Lcosθ=h is used to determine the angle, which is found to be 32.6 degrees.
  • #1
fifipaws
1
0

Homework Statement


A 15 g bullet is fired into a 2.55 kg block that is hanging by a 2.0 m long string. The speed of the bullet is 425 m/s.

Calculate the angle when the system reaches its maximum height?

Homework Equations


m1v1=(m1+m2)vf

(1/2)mv^2=mgh

The Attempt at a Solution



(0.015 kg)(425 m/s) = (0.015 kg + 2.55 kg)vf
vf=2.49 m/s

(.5)(2.565 kg)(2.49 m/s)^2=(2.565)(9.8)h
h=.316 m

Now that I have my height, I am not sure what to do. I think it should be a simple trig function at this point, but I cannot get the correct answer which is 32.6 degrees. I've also tried using L-Lcosθ=h, but have not had luck with that.

Thanks in advance.
 
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  • #2
fifipaws said:

Homework Statement


A 15 g bullet is fired into a 2.55 kg block that is hanging by a 2.0 m long string. The speed of the bullet is 425 m/s.

Calculate the angle when the system reaches its maximum height?


Homework Equations


m1v1=(m1+m2)vf

(1/2)mv^2=mgh


The Attempt at a Solution



(0.015 kg)(425 m/s) = (0.015 kg + 2.55 kg)vf
vf=2.49 m/s

(.5)(2.565 kg)(2.49 m/s)^2=(2.565)(9.8)h
h=.316 m

Now that I have my height, I am not sure what to do. I think it should be a simple trig function at this point, but I cannot get the correct answer which is 32.6 degrees. I've also tried using L-Lcosθ=h, but have not had luck with that.

Thanks in advance.

Your work is correct so far, and yes, use L-Lcosθ=h. What is your problem with it?


ehild
 

1. What is a ballistic bullet pendulum conservation of momentum problem?

A ballistic bullet pendulum conservation of momentum problem is a physics problem that involves a bullet being fired into a pendulum, causing it to swing. This problem demonstrates the principle of conservation of momentum, which states that the total momentum of a system remains constant in the absence of external forces.

2. How does the conservation of momentum apply to this problem?

In this problem, the conservation of momentum applies because the total momentum of the system (the bullet and pendulum) before and after the collision remains the same. This means that the momentum of the bullet is transferred to the pendulum, causing it to swing.

3. What information is needed to solve a ballistic bullet pendulum conservation of momentum problem?

To solve this type of problem, you will need to know the mass and velocity of the bullet, as well as the mass of the pendulum and the angle of its swing after the bullet collides with it. You may also need to know the gravitational acceleration, depending on the specific problem.

4. What are the key equations used in solving this problem?

The key equations used in solving a ballistic bullet pendulum conservation of momentum problem are the conservation of momentum equation: m1v1 = m2v2, and the equation for the pendulum's final velocity: v = √(2gh(1-cosθ)), where m1 and v1 are the mass and velocity of the bullet, m2 is the mass of the pendulum, v2 is the final velocity of the pendulum, g is the gravitational acceleration, h is the height of the pendulum, and θ is the angle of its swing.

5. What are some real-world applications of the ballistic bullet pendulum conservation of momentum problem?

The ballistic bullet pendulum conservation of momentum problem has various real-world applications, including forensic investigations to determine the speed and direction of a bullet based on its impact on an object, and in the design and testing of safety equipment for sports such as hockey and lacrosse, where the impact of a ball or puck needs to be measured and accounted for. This principle is also used in the development of airbags in cars, where the impact of a collision must be absorbed and dispersed in a controlled manner.

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