A 1.20-{\rm kg} block of wood sits at the edge of a table, 0.720 m above the floor. A 1.20×10−2-{\rm kg} bullet moving horizontally with a speed of 750 m/s embeds itself within the block
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The subject is called physics, not physic.
This is a conservation of momentum problem.
A 1.20- block of wood sits at the edge of a table, 0.720 above the floor. A 1.20×10−2- bullet moving horizontally with a speed of 750 embeds itself within the block. What horizontal distance does the block cover before hitting the ground?
To solve this problem, we need to consider the conservation of momentum and energy. First, let's analyze the initial situation before the bullet hits the block.
Given information:
Mass of the block (m_block) = 1.20 kg
Height of the table (h) = 0.720 m
Mass of the bullet (m_bullet) = 1.20 × 10^(-2) kg
Speed of the bullet (v_bullet) = 750 m/s
1. Determine the initial gravitational potential energy of the block:
Gravitational potential energy (PE) = m_block * g * h
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
PE = (1.20 kg) * (9.8 m/s^2) * (0.720 m)
2. Determine the initial momentum of the bullet:
Momentum (p_bullet) = m_bullet * v_bullet
Now, let's analyze the final situation after the bullet embeds itself within the block.
3. Determine the final velocity of the combined block and bullet:
Since the bullet embeds itself in the block, the final velocity will depend on the momentum conservation. The total momentum before and after the collision should be the same.
Initial momentum = Final momentum
m_bullet * v_bullet = (m_block + m_bullet) * v_combined
v_combined = (m_bullet * v_bullet) / (m_block + m_bullet)
4. Determine the final total energy of the system:
The total energy will be the sum of the initial gravitational potential energy and the final kinetic energy.
Total energy = PE + (1/2) * (m_block + m_bullet) * (v_combined)^2
Calculate the values and substitute them into the equations using appropriate units to find the answers for the final velocity and the total energy of the system.