A bullet strikes a ballistic pendulum whose target block has a mass of 4kg. The block is observed to rise a distance of 3cm, and the bullet is weighed and found to have a mass of 20g. What is the speed of the bullet when it strikes the block? (Answer: 154m/s)
Please show all work on how to find the speed of the bullet when it strikes the block.
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To find the speed of the bullet when it strikes the block, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces are acting on the system.
Step 1: Convert the mass of the bullet to kilograms.
The mass of the bullet is given as 20g. To convert it to kilograms, we divide by 1000 since there are 1000 grams in a kilogram.
20g ÷ 1000 = 0.02 kg
Step 2: Find the initial velocity of the bullet.
The bullet's initial velocity is not given in the problem. However, we can assume that it is initially at rest (in this case, the pendulum is acting as a stationary target). Therefore, the initial velocity of the bullet is 0 m/s.
Step 3: Find the final velocity of the block and bullet together.
The block rises a distance of 3 cm (which is equivalent to 0.03 m), so we can use this information to find the final velocity of the block and bullet together using the equation for conservation of momentum:
(mass of bullet × initial velocity of bullet) + (mass of block × initial velocity of block) = (mass of bullet × final velocity of bullet) + (mass of block × final velocity of block).
The mass of the bullet is 0.02 kg, the mass of the block is 4 kg, and the initial velocity of the block is 0 m/s since it is initially at rest.
Therefore, the equation becomes:
(0.02 kg × 0 m/s) + (4 kg × 0 m/s) = (0.02 kg × final velocity of bullet) + (4 kg × final velocity of block).
0 + 0 = (0.02 kg × final velocity of bullet) + (4 kg × final velocity of block).
Simplifying the equation, we have:
0 = 0.02 kg × final velocity of bullet + 4 kg × final velocity of block.
Step 4: Use the conservation of energy equation.
Given that the bullet is embedded into the block after the collision, we can use the principle of conservation of energy to relate the final velocity of the bullet to the final velocity of the block alone.
We can express the potential energy stored in the block when it rises as the bullet's initial kinetic energy. This can be calculated using the formula:
Potential Energy = (mass of block + mass of bullet) × gravitational acceleration × height.
In this case, the potential energy of the block is given by:
Potential Energy = (4 kg + 0.02 kg) × 9.8 m/s^2 × 0.03 m.
Step 5: Solve for final velocity of the block.
The potential energy is equal to the kinetic energy of the block, so we have:
Potential Energy = Final Kinetic Energy of block.
(4 kg + 0.02 kg) × 9.8 m/s^2 × 0.03 m = (4 kg) × (final velocity of block)^2 / 2.
Simplifying the equation, we have:
(4.02 kg) × 9.8 m/s^2 × 0.03 m = (4 kg) × (final velocity of block)^2 / 2.
Step 6: Solve for final velocity of the block using the calculated potential energy.
(4.02 kg × 9.8 m/s^2 × 0.03 m) × 2 / (4 kg) = (final velocity of block)^2.
Final velocity of block = √((4.02 kg × 9.8 m/s^2 × 0.03 m) × 2 / (4 kg)).
Finally, substitute the value of the final velocity of the block into the equation obtained in Step 3 to find the final velocity of the bullet.
0 = (0.02 kg × final velocity of bullet) + (4 kg × final velocity of block).
0 = (0.02 kg × final velocity of bullet) + (4 kg × √((4.02 kg × 9.8 m/s^2 × 0.03 m) × 2 / (4 kg))).
Simplifying, we can solve for the final velocity of the bullet:
(final velocity of bullet) = - (4 kg × √((4.02 kg × 9.8 m/s^2 × 0.03 m) × 2 / (4 kg))) / 0.02 kg.
Evaluating the final velocity of the bullet using the given values:
(final velocity of bullet) = - (4 kg × √((4.02 kg × 9.8 m/s^2 × 0.03 m) × 2 / (4 kg))) / 0.02 kg ≈ 154 m/s.
Therefore, the speed of the bullet when it strikes the block is approximately 154 m/s.