To analyze the resulting motion of the 1.2 kg mass attached to a spring after being struck by a 10-gram bullet, we can consider the situation using the principles of momentum conservation and simple harmonic motion (SHM).
-
Conservation of Momentum: When the bullet strikes the mass, we can assume that the collision is perfectly inelastic (the bullet embeds itself into the mass), which means they move together after the collision.
- Mass of the bullet, \(m_b = 0.01 , \text{kg}\)
- Mass of the mass, \(m_m = 1.2 , \text{kg}\)
Let \(v_b\) be the velocity of the bullet just before impact and \(v_f\) be the velocity of the combined system (mass + bullet) after the impact.
Using conservation of momentum: \[ m_b v_b = (m_b + m_m) v_f \] Solving for \(v_f\): \[ v_f = \frac{m_b v_b}{m_b + m_m} \]
-
Initial Conditions:
- If the spring is uncompressed and at its equilibrium position before the impact, we can analyze the motion immediately after the impact.
- The bullet imparts a momentum to the mass. The initial kinetic energy of the system can be calculated thereafter if needed.
-
Motion of the Mass-Spring System: After the bullet embeds into the mass, the system (mass + bullet) is initially moving with velocity \(v_f\).
- If we denote the spring constant as \(k\), the mass-spring system will behave as a simple harmonic oscillator after the impact. The equation of motion for a mass-spring system is given by Hooke's Law and the second law of motion.
The total mechanical energy after the collision consists of:
- Kinetic energy from the motion, which is \(\frac{1}{2}(m_b + m_m)v_f^2\).
- Potential energy stored in the spring when compressed or displaced from its equilibrium position, which is given by \(\frac{1}{2}kx^2\) where \(x\) is the displacement from the equilibrium position.
-
Resulting Motion: After the impact:
- As the system oscillates, the mass will move back and forth around the equilibrium position of the spring.
- The nature of oscillation will depend on the spring constant \(k\) and the total mass of the system \(m_b + m_m\). The angular frequency \(\omega\) of the simple harmonic motion can be calculated as: \[ \omega = \sqrt{\frac{k}{m_b + m_m}} \]
-
Damped or Undamped Motion:
- If there is any damping (e.g., air resistance, internal friction in the spring), the motion will be damped oscillation. If not, it will be undamped oscillation.
In conclusion, the resulting motion of the 1.2 kg mass attached to the spring after being struck by the 10-gram bullet will be a harmonic oscillation determined by the spring constant and the total mass of the system. The system will oscillate about the equilibrium position defined by the spring's initial rest position.
0 answers