Final answer:
The kinetic energy of a block/bullet system immediately after an inelastic collision is given by KE = \(½×m·v^2\) / (m + M), where m and v are the mass and velocity of the bullet, and M is the mass of the block. Gravity (g) is not a factor in this horizontal collision.
Step-by-step explanation:
To find the expression for the kinetic energy of the block/bullet system immediately after the collision, we first acknowledge that the collision is inelastic because internal kinetic energy is not conserved. Subsequent to an inelastic collision, the combined mass of both the bullet and block (m + M) moves with a velocity V, which is determined using the conservation of momentum. The bullet of mass m moving at velocity v collides with and embeds into a block of mass M at rest. Using conservation of momentum, the total momentum before the collision is m·v. After they collide and stick together, the new velocity (V) of the system can be calculated as:
m·v = (m + M)V
Therefore, V can be expressed as:
V = (m·v) / (m + M)
Following this, we can write the expression for the kinetic energy after the collision:
KE = \(½×(m + M)V^2\)
By substituting V from the previous equation, we get:
KE = \(½×(m + M)((m·v) / (m + M))^2\)
This simplifies to:
KE = \(½×m·v^2\) / (m + M)
We can see that the kinetic energy depends on both the mass and velocity of the bullet and the mass of the pendulum bob. This equation expresses the kinetic energy in terms of the system's defined quantities and g is not needed here as the motion is horizontal and gravity does not factor into the immediate post-collision kinetic energy.