How Fast Was the Bullet That Lodged in the Wooden Block?

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Homework Help Overview

The problem involves a 2.0-kg block of wood and a 7.0-g bullet, where the bullet lodges in the block after being shot straight up through a hole in the table. The block subsequently rises 25 cm above the tabletop, and the original poster seeks to determine the initial velocity of the bullet.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the height the block reaches and the final velocity needed to achieve that height. There are questions about how to utilize the given height in calculations and what equations might apply.

Discussion Status

Participants are exploring various approaches to relate the height of 25 cm to the initial velocity of the bullet. Some guidance has been offered regarding the use of kinematic equations and energy conservation, but no consensus has been reached on a specific method.

Contextual Notes

There is uncertainty regarding the time variable in the kinematic equations, and participants are considering the implications of using different equations that do not require time. The original poster is also navigating the constraints of the problem as part of their homework assignment.

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Homework Statement


A 2.0-kg block of wood rests on a tabletop. A 7.0-g bullet is shot straight up through a hole in the table beneath the block. The bullet lodges in the block, and the block flies 25 cm above the tabletop. How fast was the bullet going initially?

Homework Equations


initial m_bullet(initial v_bullet)+initial m_wood(initial v_wood)=(m_bullet+m_wood)Final v

The Attempt at a Solution


I know I am looking for the initial velocity of the bullet but how do I get the final velocity of the block of wood once it was hit by the bullet?
 
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Look at the information you have. Is there some piece of information that might allow you to get the final velocity?
 
the 25cm, I am just not sure how I could use it
 
creynaud said:
the 25cm, I am just not sure how I could use it
How fast would you need to throw a stone up for it to reach a height of 25cm? What equations do you know relating to constant acceleration, time, distance traveled etc?
 
s=vi(t)+1/2(a)(t)^2
what would i put for time though?
 
What is the velocity at the height where it turns around?

Edit: You can also apply energy conservation, which will be less cumbersome.
 
creynaud said:
s=vi(t)+1/2(a)(t)^2
what would i put for time though?
There are five SUVAT equations. Each involves four of the variables: distance, initial speed, final speed, time, acceleration. As you note, you do not know what time to use, and you don't need to find it. So use the one equation that does not involve time. Also see Orodruin's reply.
 

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