A 4.5 g bullet is stopped in a block of wood (m= 5.04 kg). The speed of the bullet-plus-wood combination immediately after the collision is 0.298 m/s. What is the original speed of the bullet? Answer in m/s.
I did it like this and got it wrong:
mv=MV
.298= (.0045kg)(velocity)
66.2 m/s= velocity
What did I do wrong? What do I need to do?
.0045 kg bullet at speed v
5.04 kg block at speed 0
initial momentum = .0045 v
final momentum = ( 5.04 + .0045)(.298)
so
.0045 v = 1.50
v = 334 m/s
To solve this question, you need to apply the principle of conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision.
Let's denote the mass of the bullet as m1 and the mass of the block of wood as m2. The initial velocity of the bullet is v1, and the final velocity of the bullet-plus-wood combination is v2.
First, let's express the equation for the conservation of momentum:
m1 * v1 + m2 * 0 = (m1 + m2) * v2
In this case, the block of wood is initially at rest, so its velocity (v2) is zero:
m1 * v1 + m2 * 0 = (m1 + m2) * 0
Simplifying this equation gives:
m1 * v1 = 0
However, this equation cannot give us a meaningful answer for the original speed of the bullet because it implies that the initial velocity of the bullet is zero, which is not the case.
So, the error in your calculation was assuming that the final velocity of the bullet-plus-wood combination is equal to the velocity of the bullet, which is incorrect.
To find the correct solution, you will need additional information about the collision, such as the coefficient of restitution or any other details about how the collision occurred. With this additional information, you can use the principles of conservation of momentum and kinetic energy to solve for the original speed of the bullet.