To solve this problem, we need to analyze the motion of the metre stick and the particle attached to it. The system consists of a metre stick pivoted at one end and a particle that swings down due to gravity. When the particle collides with the lower end of the stick, it will impart some energy to the stick, causing it to rise. Our goal is to find the maximum angle through which the stick will rise after the collision.
Understanding the System Dynamics
First, let’s break down the components of the system:
- The metre stick has a mass of 240 g (0.24 kg).
- The particle has a mass of 100 g (0.1 kg).
- The length of the string is 1 m, which means the particle will fall a distance of 1 m before colliding with the stick.
Energy Considerations
Initially, when the system is released, the particle has gravitational potential energy that will convert into kinetic energy as it falls. The potential energy (PE) of the particle at the top can be calculated using the formula:
PE = mgh
Where:
- m = mass of the particle = 0.1 kg
- g = acceleration due to gravity ≈ 9.81 m/s²
- h = height fallen = 1 m
Substituting the values:
PE = 0.1 kg × 9.81 m/s² × 1 m = 0.981 J
Collision and Energy Transfer
When the particle collides with the lower end of the stick, it transfers its kinetic energy to the stick. Assuming the collision is perfectly inelastic (the particle sticks to the stick), we can use the conservation of momentum to find the velocity of the combined system just after the collision.
The momentum before the collision is:
p_initial = m_particle × v_particle
Since the particle falls from rest, its velocity just before the collision can be found using the energy conservation principle:
KE = PE
Where KE is the kinetic energy of the particle just before the collision:
KE = 0.5 × m_particle × v²
Setting KE equal to the potential energy we calculated:
0.5 × 0.1 kg × v² = 0.981 J
Solving for v:
v² = (0.981 J × 2) / 0.1 kg = 19.62
v = √19.62 ≈ 4.43 m/s
Now, the momentum before the collision is:
p_initial = 0.1 kg × 4.43 m/s = 0.443 kg·m/s
After the collision, the total mass of the system is:
m_total = m_stick + m_particle = 0.24 kg + 0.1 kg = 0.34 kg
Using conservation of momentum:
p_initial = p_final
0.443 kg·m/s = 0.34 kg × v_final
Solving for v_final:
v_final = 0.443 kg·m/s / 0.34 kg ≈ 1.3 m/s
Maximum Angle Calculation
After the collision, the stick will start to rotate about the pivot. We can find the maximum angle using energy conservation again. The kinetic energy of the stick just after the collision will convert into potential energy at the maximum height:
KE_initial = PE_final
The potential energy at the maximum height can be expressed as:
PE = m_total × g × h
Here, h is the vertical height gained by the center of mass of the stick. The center of mass of the stick is at 0.5 m from the pivot. The height gained can be expressed in terms of the angle θ:
h = 0.5 × (1 - cos(θ))
Setting the kinetic energy equal to the potential energy:
0.5 × m_total × v_final² = m_total × g × h
Substituting for h:
0.5 × 0.34 kg × (1.3 m/s)² = 0.34 kg × 9.81 m/s² × (0.5 × (1 - cos(θ)))
Canceling out the mass:
0.5 × (1.3)² = 9.81 × (0.5 × (1 - cos(θ)))
Calculating the left side:
0.5 × 1.69 = 0.845
Now, simplifying the right side:
0.845 = 4.905 × (1 - cos(θ))
Solving for cos(θ):
1 - cos(θ) = 0.845 / 4.905
1 - cos(θ) ≈ 0.172
cos(θ) ≈ 0.828
Finally, calculating θ:
θ ≈ cos⁻¹(0.828) ≈ 34.5°
Final Thoughts
The maximum angle through which the stick will rise after the collision is approximately 34.5 degrees. This analysis illustrates the interplay between potential and kinetic energy in a dynamic system and how momentum conservation plays a crucial role in determining the outcomes of collisions.
