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

In summary, the problem involves a small wooden block with a mass of 0.775 kg suspended from a 1.58m long cord. A bullet with a mass of 0.0134 kg is fired at the block, becoming embedded in it and causing it to swing on the end of the cord. The tension in the cord is 4.92N when the block has risen to a height of 0.700m. Using conservation of momentum and considering the centripetal force, the final speed of the embedded objects can be found. The tension points along the string and can be visualized using a free body diagram. The initial speed of the bullet can then be calculated.
  • #1
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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  • #3
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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1. What is momentum?

Momentum is a physical quantity that describes the motion of an object. It is defined as the product of an object's mass and its velocity. In other words, it is a measure of how much "oomph" an object has as it moves.

2. What is impulse?

Impulse is a change in momentum. It is equal to the force applied to an object multiplied by the time over which the force is applied. In simpler terms, it is the amount of "push" or "pull" needed to change an object's momentum.

3. How are momentum and impulse related?

Momentum and impulse are closely related because impulse causes a change in an object's momentum. Mathematically, impulse is equal to the change in momentum, or J = Δp. This means that the greater the impulse applied to an object, the greater the change in its momentum.

4. What is a collision?

A collision occurs when two or more objects come into contact with each other and exchange energy. In the case of a wooden block being shot with a bullet, the collision would involve the bullet transferring its energy to the block, causing it to move.

5. How is momentum conserved in a collision?

In a closed system, such as the collision between a wooden block and a bullet, the total momentum before the collision is equal to the total momentum after the collision. This is known as the law of conservation of momentum. Therefore, the momentum of the bullet and the block after the collision will be equal to the momentum of the bullet before the collision.

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