A 20.0 g bullet is fired horizontally into a 85 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 156 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? (b)What was the speed of the bullet at impact with the block?
To solve this problem, we can use the law of conservation of momentum and the principle of conservation of mechanical energy. Both are fundamental principles in physics.
(a) To find the velocity of the block once the bullet is embedded, we first need to find the initial velocity of the block-bullet system before the collision. We can assume that the initial velocity of the bullet is "V" (in m/s) and the mass of the block is "m." The bullet is embedded in the block, so the total mass of the system after the collision becomes (m + 0.020 kg).
Using the conservation of momentum, we can write:
Initial momentum = Final momentum
(momentum of bullet) + (momentum of block) = (momentum of block-bullet system after collision)
(mV) + 0 = (m + 0.020 kg) * Vf
Simplifying this equation, we get:
mV = (m + 0.020 kg) * Vf
Now, to find the velocity of the block (Vf), we also need to consider that some of the initial kinetic energy of the bullet is converted into potential energy of the compressed spring. This energy is given by the equation:
Potential Energy = (1/2) * k * x^2
Where "k" is the spring constant and "x" is the displacement of the spring from its equilibrium position. In this case, the maximum compression of the spring is given as 0.013 m.
The final kinetic energy of the block-bullet system is the same as the initial kinetic energy of the bullet. So, we can write:
Initial kinetic energy of bullet = Final kinetic energy of block-bullet system
(1/2) * m * V^2 = (1/2) * (m + 0.020 kg) * Vf^2
Now, we have two equations:
mV = (m + 0.020 kg) * Vf (Equation 1)
(1/2) * m * V^2 = (1/2) * (m + 0.020 kg) * Vf^2 (Equation 2)
Using Equation 2, we can solve for Vf:
(m + 0.020 kg) * Vf^2 = (m * V^2) / (1 + 0.020 kg / m)
Vf^2 = (V^2) / (1 + 0.020 kg / m)
Vf = sqrt[(V^2) / (1 + 0.020 kg / m)]
Substituting this value of Vf into Equation 1, we can solve for the velocity of the block (Vf):
mV = (m + 0.020 kg) * sqrt[(V^2) / (1 + 0.020 kg / m)]
(b) To find the speed of the bullet at impact with the block, we can use the formula:
Kinetic energy = (1/2) * m * V^2
where "m" is the mass of the bullet and "V" is the speed of the bullet. Given the mass of the bullet (20.0 g = 0.020 kg), we can rearrange the formula to solve for "V":
V = sqrt(2 * kinetic energy / m)
Substituting the values given in the problem, we get:
V = sqrt(2 * (1/2) * 0.020 kg * (V^2))
Simplifying, we have:
V = sqrt(V^2)
Taking the square root of both sides, we find:
V = |V|
Since speed is always positive, the speed of the bullet at impact with the block is simply equal to the magnitude of the bullet's velocity.