Math Problem Statement

In a ballistic pendulum experiment, a dart with a mass of 0.050 kg is shot with an initial speed of 28 m/s at a target block that is initially at rest. The block has a mass of 3.0 kg. After the dart embeds itself in the block, they swing upward together. Determine the maximum height reached by the dart-block system.

Solution

To determine the maximum height reached by the dart-block system, we’ll need to apply two main principles of physics: conservation of momentum for the collision phase and conservation of energy for the swinging motion after the collision.

Step-by-Step Solution

  1. Conservation of Momentum for the Collision: When the dart embeds itself into the block, the momentum of the dart-block system is conserved because there are no external horizontal forces acting on it.

    Let:

    • m1=0.050kgm_1 = 0.050 \, \text{kg} be the mass of the dart.
    • v1=28m/sv_1 = 28 \, \text{m/s} be the initial velocity of the dart.
    • m2=3.0kgm_2 = 3.0 \, \text{kg} be the mass of the block.
    • v2=0m/sv_2 = 0 \, \text{m/s} be the initial velocity of the block.

    After the dart embeds into the block, the total mass becomes: M=m1+m2=0.050+3.0=3.05kgM = m_1 + m_2 = 0.050 + 3.0 = 3.05 \, \text{kg}

    Using the conservation of momentum, we find the velocity vfv_f of the combined dart-block system just after the collision: m1v1+m2v2=(m1+m2)vfm_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f Substituting the known values: (0.050)(28)+(3.0)(0)=(3.05)vf(0.050)(28) + (3.0)(0) = (3.05) v_f 1.4=3.05vf1.4 = 3.05 v_f vf=1.43.050.459m/sv_f = \frac{1.4}{3.05} \approx 0.459 \, \text{m/s}

  2. Conservation of Energy for the Swing: After the collision, the dart-block system swings upward. At the highest point of the swing, all the initial kinetic energy (from the velocity vfv_f just after the collision) is converted into potential energy.

    The kinetic energy right after the collision is: KE=12Mvf2\text{KE} = \frac{1}{2} M v_f^2 Substituting the values: KE=12(3.05)(0.459)2\text{KE} = \frac{1}{2} (3.05) (0.459)^2 KE0.321J\text{KE} \approx 0.321 \, \text{J}

    At the maximum height hh, all this kinetic energy will convert into gravitational potential energy, which is given by: PE=Mgh\text{PE} = Mgh Setting the initial kinetic energy equal to the potential energy at maximum height: 0.321=(3.05)(9.8)h0.321 = (3.05)(9.8) h Solving for hh: h=0.3213.05×9.8h = \frac{0.321}{3.05 \times 9.8} h0.0107mh \approx 0.0107 \, \text{m}

Final Answer

The maximum height reached by the dart-block system is approximately: h0.0107mor1.07cmh \approx 0.0107 \, \text{m} \quad \text{or} \quad 1.07 \, \text{cm}

Would you like further details on this problem or have any specific questions?


Related Questions

  1. What would happen to the maximum height if the dart’s mass were doubled?
  2. How would the result change if there was friction during the swing?
  3. What is the importance of using conservation of momentum in this problem?
  4. How would a different initial velocity for the dart affect the outcome?
  5. How would air resistance affect the final height?

Tip:

When solving physics problems involving collisions and swings, remember to separate the stages of the motion (collision and swing) and apply the appropriate conservation law for each stage.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Conservation of Momentum
Conservation of Energy
Kinematics
Work and Energy

Formulas

m1 * v1 + m2 * v2 = (m1 + m2) * vf
KE = (1/2) * M * vf^2
PE = M * g * h

Theorems

Conservation of Momentum
Conservation of Energy

Suitable Grade Level

Grades 11-12