A 16.0 g bullet is fired horizontally into a 87 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 168 N/m. The bullet becomes embedded in the block. The bullet block system compresses the spring by a maximum of 1.30 cm. (a) What is the velocity of the block once the bullet is imbedded?
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To find the velocity of the block once the bullet is embedded, we can use the principle of conservation of momentum.
The equation for conservation of momentum is:
m_bullet * v_bullet + m_block * v_block = (m_bullet + m_block) * vf
where:
m_bullet is the mass of the bullet (16.0 g)
v_bullet is the velocity of the bullet
m_block is the mass of the block (87 g)
v_block is the initial velocity of the block (which is zero as it is initially at rest)
vf is the final velocity of the bullet-block system after the bullet is embedded
We need to find vf (final velocity of the bullet-block system).
First, let's convert the masses to kilograms:
m_bullet = 16.0 g = 0.016 kg
m_block = 87 g = 0.087 kg
Now, since the block is initially at rest, the momentum of the block is zero. Therefore, the equation becomes:
m_bullet * v_bullet = (m_bullet + m_block) * vf
Substituting the given values, we have:
0.016 kg * v_bullet = (0.016 kg + 0.087 kg) * vf
Simplifying further, we get:
0.016 kg * v_bullet = 0.103 kg * vf
Now we can solve for vf. Rearranging the equation, we have:
vf = (0.016 kg * v_bullet) / 0.103 kg
Now, we need to relate the compression of the spring to the final velocity of the bullet-block system.
The potential energy stored in a spring is given by the equation:
U = (1/2) * k * x^2
where:
U is the potential energy stored in the spring
k is the spring constant (168 N/m)
x is the displacement or compression of the spring (1.30 cm = 0.013 m)
The potential energy stored in the spring is equal to the kinetic energy of the bullet-block system:
U = (1/2) * (m_bullet + m_block) * vf^2
Substituting the given values, we have:
(1/2) * (0.016 kg + 0.087 kg) * vf^2 = (1/2) * 168 N/m * (0.013 m)^2
Now we can solve for vf^2. Simplifying further, we have:
(0.103 kg) * vf^2 = (1/2) * 168 N/m * (0.013 m)^2
vf^2 = [(1/2) * 168 N/m * (0.013 m)^2] / (0.103 kg)
vf^2 = 0.01547 m^2/s^2
Now, taking the square root of both sides, we find:
vf ≈ 0.124 m/s
Therefore, the velocity of the block once the bullet is embedded is approximately 0.124 m/s.