A 1.75-kg wooden block rests on a table over a large hole as in the figure below. A 4.60-g bullet with an initial velocity vi is fired upward into the bottom of the block and remains in the block after the collision. The block and bullet rise to a maximum height of 24.0 cm.
how to start
To find the initial velocity (vi) of the bullet, we can use the principle of conservation of energy and momentum.
First, let's calculate the potential energy when the block and bullet reach the maximum height. The potential energy (PE) is given by:
PE = mgh
Where:
m = mass (total mass of the block and bullet)
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height (24.0 cm converted to meters)
The mass of the block (m_block) is 1.75 kg, and the mass of the bullet (m_bullet) is 4.60 g, which is 0.0046 kg. Therefore, the total mass (m) is:
m = m_block + m_bullet
Plugging in the values, we have:
m = 1.75 kg + 0.0046 kg
Next, convert the height from centimeters to meters:
h = 24.0 cm / 100
Now, we can calculate the potential energy:
PE = (1.75 kg + 0.0046 kg) * 9.8 m/s^2 * 0.24 m
Solving this equation will give you the potential energy when the block and bullet reach the maximum height.
Once we have the potential energy, we can use the principle of conservation of energy. Before the collision, the system had kinetic energy due to the bullet's initial velocity (KE_bullet), and after the collision, the system had potential energy at the maximum height (PE).
The bullet's kinetic energy can be expressed as:
KE_bullet = (1/2) * m_bullet * vi^2
Since the bullet is lodged in the block after the collision, both the block and bullet are moving together. Therefore, the mass considered for the kinetic energy will be the total mass (m).
The total kinetic energy after the collision can be expressed as:
KE = (1/2) * m * v^2
Since the system is at rest initially, the total initial kinetic energy is zero (KE_initial = 0).
By applying the principle of conservation of energy, we can equate the initial kinetic energy to the final kinetic energy plus the potential energy:
KE_initial = KE + PE
0 = (1/2) * m * v^2 + (1.75 kg + 0.0046 kg) * 9.8 m/s^2 * 0.24 m
Rearranging the equation and solving for v (velocity), we can find the initial velocity (vi) of the bullet:
(1/2) * m * v^2 = - (1.75 kg + 0.0046 kg) * 9.8 m/s^2 * 0.24 m
vi = sqrt((- 2 * g * h * (m_block + m_bullet)) / m)
Plugging in the values of g, h, m_block, and m_bullet will give you the initial velocity (vi) of the bullet.