Initial Speed of 5.5g Bullet Fired into 22.6g Wood Block

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A 5.5 g bullet is fired into a 22.6 g wood block resting on a 1.5 m tall post, and after the collision, they land 2.5 m from the base. To find the bullet's initial speed, first calculate the time it takes for the block to fall 1.5 m using the equation d = V_it + 0.5at^2. With the time determined, the horizontal velocity can be found using V = d/t. The conservation of momentum can then be applied to relate the bullet's initial speed to the combined velocity of the bullet and block after the collision. This problem involves multiple steps and requires careful application of physics principles.
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A 5.5 g bullet is fired into a block of wood with a mass of 22.6g. The wood block is intially at rest on a 1.5m tall post. After the collision, the wood block and bullet land 2.5m from the base of the post. Find the initial speed of the bullet.
 
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I need help starting it! I'm sort of confused. Thanks for your time.
 
Originally posted by Addison
A 5.5 g bullet is fired into a block of wood with a mass of 22.6g. The wood block is intially at rest on a 1.5m tall post. After the collision, the wood block and bullet land 2.5m from the base of the post. Find the initial speed of the bullet.

Ah the great many many many step question.

First find the time it took to fall the 1.5m using this equation

d = V_it + \frac{1}{2}at^2

Since you know the time and the horizontal distance, find the horizontal velocity using this formula

V = \frac{d}{t}

Now that you know the horizontal velocity of the block and and the bullet, you can also find and compare the inertia. Subscript bu will be bullet and bl will be block.

(V_b_u) (m_b_u) = (V_b_u_+_b_l) (m_b_u_+_b_l)


I hope I didn't leave anything out.
 
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