Bullet Through Ballistic Pendulum- Please help me

[shigami117]

1. A 9.05- g bullet from a 9-mm pistol has a velocity of 331.0 m/s. It strikes the 0.705- kg block of a ballistic pendulum and passes completely through the block. If the block rises through a distance h = 19.47 cm, what was the velocity of the bullet as it emerged from the block? i know that this is supposed to be done using conservation of energy and momentum. but i can't get it

 

[mgb_phys]

It's still a conservation of energy.

Initial kinetic energy of bullet + initial potential energy of block (which is zero) -> final ke of bullet + final potential energy of block.

 

[thomate1]

Hi there,

I understand you are struggling with solving this problem using conservation of energy and momentum. Let me walk you through the steps to help you understand the solution.

First, let's define our variables. We have a bullet with a mass of 9.05 g (0.00905 kg) and a velocity of 331.0 m/s. It strikes a block with a mass of 0.705 kg and causes it to rise through a distance of 19.47 cm (0.1947 m). We want to find the velocity of the bullet as it emerges from the block, let's call it v'.

Now, let's apply conservation of momentum. We know that momentum is conserved in a closed system, so the total momentum before the collision must equal the total momentum after the collision. In this case, the bullet is the only object moving before and after the collision, so we can write:

m1v1 = m1v1' + m2v2'

Where m1 is the mass of the bullet, v1 is its initial velocity, m2 is the mass of the block, v1' is the velocity of the bullet after the collision, and v2' is the velocity of the block after the collision.

Substituting our values, we get:

(0.00905 kg)(331.0 m/s) = (0.00905 kg)(v') + (0.705 kg)(v2')

Now, let's apply conservation of energy. We know that the total energy before the collision (kinetic energy of the bullet) must equal the total energy after the collision (kinetic energy of the bullet and block). So we can write:

(1/2)m1v1^2 = (1/2)m1v1'^2 + (1/2)m2v2'^2

Substituting our values, we get:

(1/2)(0.00905 kg)(331.0 m/s)^2 = (1/2)(0.00905 kg)(v')^2 + (1/2)(0.705 kg)(v2')^2

Now, we have two equations with two unknowns (v' and v2'). We can solve for one of the variables and then substitute it into the other equation to solve for the other variable.

Let's solve for v2' first. Rearranging the momentum equation, we get: