Momentum, Impulse, and Collisions: Wooden Block Shot with a bullet

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SUMMARY

The discussion centers on a physics problem involving a wooden block with a mass of 0.775 kg and a bullet with a mass of 0.0134 kg, where the bullet is fired at the block, resulting in a perfect inelastic collision. The final velocity of the combined object after the collision is calculated using the conservation of momentum formula: m1v1 + m2v2 = (m1 + m2)vf. The tension in the cord when the block rises to a height of 0.700 m is given as 4.92 N, and participants discuss how to visualize the forces acting on the block and bullet system, particularly in relation to centripetal force and free body diagrams.

PREREQUISITES
  • Understanding of perfect inelastic collisions
  • Knowledge of conservation of momentum principles
  • Familiarity with centripetal force and acceleration concepts
  • Ability to draw and interpret free body diagrams
NEXT STEPS
  • Calculate the initial speed V0 of the bullet using the derived equations
  • Explore the implications of unbalanced forces in momentum conservation
  • Learn how to construct and analyze free body diagrams in collision scenarios
  • Study the relationship between tension in strings and the forces acting on pendulum-like systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, collision theory, and forces in motion. This discussion is beneficial for anyone looking to deepen their understanding of momentum and tension in physical systems.

ahorowitz1
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1.So the problem given is the following, "A small wooden block with mass 0.775 kg is suspended from the lower end of a light cord that is 1.58m long. The block is initially at rest. A bullet with mass 0.0134 kg is fired at the block with a horizontal velocity[UP][/UP] v0. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.700 m , the tension in the cord is 4.92N."


2. Right now the I can figure out a few equations from this. I know that because the bullet becomes embedded in the wooden block, it can be considered a perfect inelastic collision. Thus m1v1 + m2v2 = (m1 + m2)vf. For this problem, I'm going to call the bullet m1 and the wooden block m2. Thus, I know that from this I can solve for the final velocity of the two embedded object. Also, I figure that because the block is on a string, and it is moving, that centripetal force/ acceleration must come into play. Therefore, should I use the equation FC=mac=T - mg?

3. vf=(0.0134kg * v0+ 0.775 kg * 0 m/s) / (0.0134 kg + 0.775 kg) = 0.016996 v0. However, I'm confused as to how to progress from here? I know that the tension when the object (combined) has reached 0.700 m is equal to 4.92 N, but I'm having difficult picturing this. Does this tension value represent the tension when the 1.58 m string is facing right, forming a triangle of sorts where 1.58 is the hypotenuse and the y-component is 0.88m? Please let me know how to progress from here, or at least if you know how I should be visualizing this that would be so helpful! Thanks!

EDIT: The question is what is the initial speed V0 of the bullet?
 
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Welcome to PF;
rude man is correct - you have not provided a complete problem statement, just a bunch of facts. I'm guessing you need to find ##v_0##.

You did ask a question of sorts:
Does this tension value represent the tension when the 1.58 m string is facing right, forming a triangle of sorts where 1.58 is the hypotenuse and the y-component is 0.88m? Please let me know how to progress from here, or at least if you know how I should be visualizing this that would be so helpful! Thanks!
The tension points along the string and is the same as how you'd normally view the tension.

You should be able to sketch the final position - you know the length of the cord and you know the vertical height is has swung to.
If you draw a free body diagram, you should be able to figure out the centripetal force and thus the final speed.

Take care using conservation of momentum when there is an unbalanced force.
 
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