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A 20-g bullet moving horizontally with a speed of 4 km/s strikes and passes through a 2.0-kg block initially at rest on a horizontal frictionless surface. the bullet emerges from the block with a speed of 1.5 km/s. what is the kinetic energy of the block immediately after the bullet emerges?

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Final answer:

Using conservation of momentum, we calculate the block's speed after the bullet emerges to be 25 m/s. The kinetic energy of the block is then found to be 625 Joules.

Step-by-step explanation:

The question involves a collision between a bullet and a block, which is a classic problem in the subject of physics, specifically in the conservation of momentum and kinetic energy in a system. When the bullet, with a mass of 0.02 kg and moving at a speed of 4000 m/s, emerges from the block of 2.0 kg at a speed of 1500 m/s, we can calculate the final kinetic energy of the block using the principle of conservation of momentum.

Step 1: Calculate the initial and final momentum of the bullet.

Initial momentum of the bullet = mass of the bullet × initial speed of the bullet
= 0.02 kg × 4000 m/s = 80 kg·m/s

Final momentum of the bullet = mass of the bullet × final speed of the bullet
= 0.02 kg × 1500 m/s = 30 kg·m/s

Since there are no external forces acting on the system (bullet + block), the initial momentum of the system must equal the final momentum.

Step 2: Calculate the velocity of the block after the bullet emerges using the conservation of momentum.

Initial momentum of the system = final momentum of the system

80 kg·m/s = (mass of the block × velocity of the block after impact) + 30 kg·m/s

Velocity of block after impact = (80 kg·m/s - 30 kg·m/s) / 2.0 kg = 25 m/s

Step 3: Calculate the kinetic energy of the block after the bullet emerges.

Kinetic energy of the block = 0.5 × mass of the block × (velocity of the block after impact)^2

= 0.5 × 2.0 kg × (25 m/s)^2 = 625 J

Therefore, the kinetic energy of the block immediately after the bullet emerges is 625 Joules.

User Jan Tumanov
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