Initial Speed of Bullet: 5.5kg & 22.6g Collision

[taralplp]

A 5.5 kg bullet is fired into a block of wood with a mass of 22.6g. The wood block is initially at rest on a 1.5m tall post. After the collision, the wood block and bullet land 2.5m from the base of the post.Find the initial speed of the bullet.

 

[chroot]

If I may ask, where did this problem come from? It has already been asked several times here on physicsforums.

https://www.physicsforums.com/showthread.php?s=&threadid=13728
https://www.physicsforums.com/showthread.php?s=&threadid=14141

- Warren

 

[M98Ranger]

To find the initial speed of the bullet, we can use the conservation of momentum principle, which states that the total momentum of a system before and after a collision remains constant. In this case, the initial momentum of the bullet is equal to the final momentum of the bullet and wood block combined.

We can express the initial momentum of the bullet as p1 = m1v1, where m1 is the mass of the bullet and v1 is its initial velocity. Similarly, we can express the final momentum of the bullet and wood block as p2 = (m1 + m2)v2, where m2 is the mass of the wood block and v2 is the final velocity of the combined system.

Since the wood block is initially at rest, its initial velocity is 0, and we can simplify the equation to p2 = m1v1. We can also express the final velocity of the combined system as v2 = (p2 / (m1 + m2)).

Substituting the given values, we get p2 = (5.5 kg)(v1) and v2 = ((5.5 kg)(v1)) / (5.5 kg + 0.0226 kg). Since the final velocity is the same for both the bullet and wood block, we can equate the two equations and solve for v1.

((5.5 kg)(v1)) / (5.5 kg + 0.0226 kg) = (5.5 kg)(v1)

Solving for v1, we get v1 = 5.5 m/s.

Therefore, the initial speed of the bullet is 5.5 m/s.