 Physics
28 November, 18:25

# A 0.050 kg bullet strikes a 5.0 kg stationary wooden block and embeds itself in the block. The block and the bullet fly off together at 9.0 m/s. What was the original velocity of the bullet?

+3
Answers (1)
1. 28 November, 19:47
0
Answer: 909 m/s

Explanation:

Given

Mass of the bullet, m1 = 0.05 kg

Mass of the wooden block, m2 = 5 kg

Final velocities of the block and bullet, v = 9 m/s

Initial velocity of the bullet v1 = ? m/s

From the question, we would notice that there is just an object (i. e the bullet) moving before the collision. Also, even after the collision between the bullet and wood, the bullet and the wood would move as one object. Thus, we would use the conservation of momentum to solve

m1v1 = (m1 + m2) v, on substituting, we have

0.05 * v1 = (0.05 + 5) * 9

0.05 * v1 = 5.05 * 9

0.05 * v1 = 45.45

v1 = 45.45 / 0.05

v1 = 909 m/s

Thus, the original velocity of the bullet was 909 m/s
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A 0.050 kg bullet strikes a 5.0 kg stationary wooden block and embeds itself in the block. The block and the bullet fly off together at 9.0 ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.