A 7.65- g bullet from a 9-mm pistol has a velocity of 329.0 m/s. It strikes the 0.625- kg block of a ballistic pendulum and passes completely through the block. If the block rises through a distance h = 29.75 cm, what was the velocity of the bullet as it emerged from the block?
To find the velocity of the bullet as it emerged from the block, we can use the principle of conservation of momentum and energy.
First, let's calculate the initial momentum of the bullet before it strikes the block:
Momentum (p) = mass (m) * velocity (v)
Initial momentum of bullet = (7.65 g) * (329.0 m/s) = 2.51685 kg·m/s
According to the principle of conservation of momentum, the total momentum before and after the collision should be the same. Since the bullet passes completely through the block, the entire momentum of the bullet is transferred to the block.
The block, with mass (M) = 0.625 kg, rises through a distance (h) = 29.75 cm. To find the velocity of the block, we can use the principle of conservation of energy. The potential energy gained by the block is equal to the kinetic energy lost by the bullet.
The potential energy gained by the block is given by:
Potential Energy (PE) = mass (M) * gravity (g) * height (h)
PE = (0.625 kg) * (9.8 m/s^2) * (0.2975 m) = 1.78977 J
The kinetic energy lost by the bullet is given by:
Kinetic Energy (KE) = (1/2) * mass (m) * velocity^2
KE = (1/2) * (0.00765 kg) * (329.0 m/s)^2 = 417.83982 J
As per the principle of conservation of energy, the kinetic energy lost by the bullet is equal to the potential energy gained by the block:
KE = PE
417.83982 J = 1.78977 J
Now, let's calculate the velocity of the block as it rises:
Kinetic Energy (KE) = (1/2) * mass (M) * velocity^2
1.78977 J = (1/2) * (0.625 kg) * velocity^2
3.57954 J/kg = (0.3125 kg) * velocity^2
Dividing both sides by 0.3125 kg:
11.45134 J/kg = velocity^2
Taking the square root of both sides:
Velocity = √(11.45134 J/kg) = 3.38 m/s
Therefore, the velocity of the bullet as it emerged from the block is approximately 3.38 m/s.
To solve this problem, we can use the principle of conservation of linear momentum.
The bullet is initially moving with a velocity of 329.0 m/s. It strikes the block and passes completely through, causing the block to rise.
We need to find the velocity of the bullet as it emerges from the block. Let's assume this velocity is v.
The momentum before the collision is equal to the momentum after the collision:
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
Where:
m1 is the mass of the bullet (7.65 g = 0.00765 kg)
v1 is the initial velocity of the bullet (329.0 m/s)
m2 is the mass of the block (0.625 kg)
v2 is the initial velocity of the block (0 m/s since it is at rest)
m1 is the mass of the bullet after emerging from the block (0.00765 kg)
v1' is the final velocity of the bullet after emerging from the block
m2 is the mass of the block after the collision (0.625 kg)
v2' is the final velocity of the block after the collision (unknown)
The bullet passes completely through the block, so its mass and velocity do not change. Therefore, m1' = m1 and v1' = v1.
The equation simplifies to:
(m1 * v1) + (m2 * v2) = (m1' * v1') + (m2 * v2')
(0.00765 kg * 329.0 m/s) + (0.625 kg * 0 m/s) = (0.00765 kg * v1') + (0.625 kg * v2')
Simplifying further:
0.00765 kg * 329.0 m/s = 0.00765 kg * v1' + 0.625 kg * v2'
Multiplying:
2.50185 kg·m/s = 0.00765 kg * v1' + 0.625 kg * v2'
Since the bullet passes through the block, its final velocity v1' is the same as its initial velocity v1:
2.50185 kg·m/s = 0.00765 kg * 329.0 m/s + 0.625 kg * v2'
Simplifying:
v2' = (2.50185 kg·m/s - 0.00765 kg * 329.0 m/s) / 0.625 kg
Calculating:
v2' = (2.50185 kg·m/s - 2.52585 kg·m/s) / 0.625 kg
v2' = (-0.024 kg·m/s) / 0.625 kg
v2' = -0.0384 m/s
The negative sign indicates that the block moves in the opposite direction of the bullet.
Therefore, the velocity of the bullet as it emerges from the block is approximately -0.0384 m/s.