A bullet (15 g) is shot horizontally into a block of wood (600 g), which is hanging in the end of a 300 cm weightless rope. The collision is fully inelastic. Calculate the velocity of the
bullet, if the rope makes an angle of 45 degree to the vertical when it is at the highest position.
To calculate the velocity of the bullet, we can use the principle of conservation of linear momentum. In an inelastic collision, the total momentum before the collision is equal to the total momentum after the collision.
Let's denote the original velocity of the bullet as v and the velocity of the block of wood (which is initially at rest) as 0. After the collision, they both move together with the same velocity, which we'll call V.
The momentum before the collision is given by the sum of the momenta of the bullet and the wood block:
momentum before = momentum of bullet + momentum of block of wood
mv = (m_b * v) + (m_w * 0) (since the initial velocity of the wood block is 0)
Since the collision is fully inelastic, the two masses stick together and move as one. Therefore, the total mass after the collision is the sum of the masses of the bullet and the wood block:
total mass after = m_b + m_w
The momentum after the collision is given by the product of the total mass and the final velocity:
momentum after = (m_b + m_w) * V
Setting the conservation of momentum equation equal to each other, we have:
mv = (m_b + m_w) * V
Now we can solve for V by substituting the values given in the problem:
v * m_b = (m_b + m_w) * V
Since the bullet and the wood block have different masses, we need to calculate the total mass using the given values:
total mass after = 15 g + 600 g = 615 g = 0.615 kg
Substituting the values into the equation, we have:
v * 15 g = (15 g + 600 g) * V
Simplifying the equation by canceling out the mass unit of grams:
v * 0.015 kg = (0.015 kg + 0.615 kg) * V
v * 0.015 kg = 0.63 kg * V
Dividing both sides of the equation by 0.015 kg:
v = (0.63 kg * V) / 0.015 kg
Now, we need to find the value of V. To do that, we can use the information about the angle the rope makes with the vertical.
When the rope is at its highest position, the tension in the rope is equal to the weight of the block.
tension = weight of block
The weight of the block is given by the product of its mass and the acceleration due to gravity (g):
weight of block = m_w * g
Substituting the given values, we have:
weight of block = 600 g * 9.8 m/s^2
Next, we can resolve the weight of the block into its vertical and horizontal components. The vertical component is equal to the tension in the rope:
vertical component of weight = tension
The vertical component of the weight is given by:
vertical component = weight of block * sin(angle)
Substituting the given angle of 45 degrees (converted to radians) and the weight of the block, we have:
vertical component = (600 g * 9.8 m/s^2) * sin(45°)
Finally, we can relate the vertical component of the weight to the net vertical force acting on the block, which is given by the product of the total mass and the acceleration in the vertical direction:
vertical component = (m_b + m_w) * g * cos(angle)
Substituting the given angle of 45 degrees (converted to radians) and the total mass, we have:
vertical component = (0.615 kg) * 9.8 m/s^2 * cos(45°)
Now, we can find V by equating the two expressions for the vertical component:
(600 g * 9.8 m/s^2) * sin(45°) = (0.615 kg) * 9.8 m/s^2 * cos(45°) * V
Simplifying the equation by canceling out the unit of grams:
(600 * 9.8 m/s^2) * sin(45°) = (0.615 kg) * 9.8 m/s^2 * cos(45°) * V
Dividing both sides of the equation by (0.615 kg) * 9.8 m/s^2 * cos(45°), we have:
V = [(600 * 9.8 m/s^2) * sin(45°)] / [(0.615 kg) * 9.8 m/s^2 * cos(45°)]
Simplifying further, we have:
V = sin(45°) / (0.615 kg * cos(45°))
V = (1 / sqrt(2)) / (0.615 kg * 1 / sqrt(2))
V = 1 / 0.615 kg
Therefore, V = 1.626 m/s
Now, substituting the value of V back into the equation for v, we have:
v = (0.63 kg * 1.626 m/s) / 0.015 kg
Simplifying, we find:
v = 68.36 m/s
Hence, the velocity of the bullet is approximately 68.36 m/s.