Question

A bullet is fired vertically into a 1.40 kg block of wood at rest directly above it. if the bullet has a mass of 29.0 g and a speed of 510 m/s, how high will the block rise after the bullet becomes embedded in it?

Answer 1

To determine how high the block will rise after a bullet becomes embedded in it, we use the conservation of momentum to find the block-bullet system's velocity right after the collision, then apply conservation of energy to calculate the maximum height the system attains.

Step-by-step explanation:

The subject of this question is Physics, specifically involving concepts from mechanics and conservation of energy. To find the height to which the block will rise after the bullet becomes embedded in it, we can use the principle of conservation of momentum for the collision and the principle of conservation of mechanical energy for the subsequent motion of the block-bullet system.

First, we calculate the initial momentum of the bullet before the collision and set it equal to the momentum of the combined block-bullet system immediately after the collision, as no external horizontal forces are acting on the system. The block's initial vertical velocity can thus be determined.

Then, we use the conservation of mechanical energy, converting the kinetic energy of the upward-moving block-bullet system to gravitational potential energy at the peak of its rise, to find the maximum height attained. The formula used is:

• Kinetic Energy Initial = Potential Energy Final
• (1/2)(mass of block + mass of bullet)(velocity after collision)^2 = (mass of block + mass of bullet)g(height)

Where g is the acceleration due to gravity (9.81 m/s2). By solving for height, we can find the answer to the question.

Answer 2

5.47 m
The bullet undergoes a non-elastic collision with the block of wood and momentum is conserved. The initial momentum is 0.029 kg * 510 m/s = 14.79 kg*m/s. The combined mass of the block and bullet is 1.40 kg * 0.029 kg = 1.429 kg. Since momentum is conserved, the velocity of both combined will then be 14.79 kg*m/s / 1.429 kg = 10.34989503 m/s.
With a local gravitational acceleration of 9.8 m/s^2, it will take 10.34989503 m/s / 9.8 m/s^2 = 1.056111738 s for their upward velocity to drop to 0, just prior to descending.
The equation for distance under constant acceleration is
d = 0.5 A T^2
so
d = 0.5 * 9.8 m/s^2 * (1.056111738 s)^2
d = 4.9 m/s^2 * 1.115372003 s^2
d = 5.465322814 m
Rounding to 3 significant figures gives a height of 5.47 meters.