<p>Question: A 2.00 g bullet hits and becomes embedded in a 5.00 kg wood block w

Question

<p>Question: A 2.00 g bullet hits and becomes embedded in a 5.00 kg wood block which is hanging from a 1.40 m long string. This causes the block to swing through an arc of 4.50 degrees. What was the speed of the bullet before it hit the block?</p>
<p>I used the equation 1/2mv^2 = mgh to start off with but got stuck when I was trying to find the velocity from the height and so continue with the problem.I think the answer might be 728 m/s but am not sure.Any help would be much appreciated! Thanks :)</p>

Explanation / Answer

Yeah, you did it right.

First, find the height that the block rose, using the angle of the arc and cos().

Once you have "h", using (1/2)mv2 = mgh, find "v", which is the velocity of the wood block and the bullet as soon as the bullet hits the block and becomes one mass with it.

Finally, use the law of conservation of momentum, m1v1 + m2v2 = (m1 + m2)v, to find the initial velocity of the bullet, since you know the masses of both and the initial velocity of the block.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.