How Is Momentum Conserved in a Bullet and Ballistic Pendulum Collision?

[stupif]

1. a 7g bullet is fired into a 1.5kg ballistic pendulum. the bullet emerges from the block with a speed of 200m/s, and the block rises to a maximun height of 12cm. find the initial speed of the bullet and the kinetic energy lost of the bullet.
answer: 528m/s, 835.7j
2. mu+ mu = mv + mv
0.007u + 1.5(0)= 0.007(200) + 1.5kg(v)
how to know the final velocity of ballistic pendulum?? help me...please...
thank you

 

[rock.freak667]

The pendulum's max height is 12cm, so its KE at the collision = PE at 12cm.

so 0.5mv²=mg(0.12).

 

[stupif]

i got it...thank you very much~

 

[stupif]

https://www.physicsforums.com/showthread.php?t=519410

can you help me this question?

 

[rwisz]

Hello,

I am happy to help you understand the concept of momentum and collision. In the first scenario, we have a bullet that is fired into a ballistic pendulum. The bullet has a mass of 7g and a speed of 200m/s. Upon collision with the pendulum, it emerges with a speed of 528m/s and the pendulum rises to a height of 12cm. To find the initial speed of the bullet, we can use the principle of conservation of momentum. This states that the total momentum before the collision is equal to the total momentum after the collision. In this case, we have:

m1v1 = m2v2

Where m1 is the mass of the bullet, v1 is its initial velocity, m2 is the mass of the pendulum and v2 is its final velocity.

Plugging in the values, we get:

0.007kg x 200m/s = 1.5kg x v2

Solving for v2, we get:

v2 = 528m/s

This means that the final velocity of the pendulum after the collision is 528m/s.

To find the kinetic energy lost by the bullet, we can use the formula:

KE = 1/2mv^2

Where m is the mass of the bullet and v is its initial velocity.

Plugging in the values, we get:

KE = 1/2 x 0.007kg x (200m/s)^2 = 835.7 joules

This is the amount of kinetic energy lost by the bullet during the collision.

In the second scenario, we have two particles with masses mu and mu, initially moving towards each other with speeds of 0.007u and 0 m/s respectively. After the collision, one particle has a mass of 0.007u and a speed of 200m/s, while the other has a mass of 1.5kg and an unknown final velocity.

To find the final velocity of the second particle, we can again use the principle of conservation of momentum. This gives us the equation:

m1v1 + m2v2 = m1v1' + m2v2'

Where m1 and m2 are the masses of the two particles, v1 and v2 are their initial velocities, and v1' and v2' are their final velocities.

Plugging in the values