Math Problem Statement

A body of mass m=1 kgm = 1 , \text{kg}m=1kg and a body of mass mz=4 kgm_z = 4 , \text{kg}mz​=4kg are on an inclined plane at heights h1=2 mh_1 = 2 , \text{m}h1​=2m and hz=1 mh_z = 1 , \text{m}hz​=1m, respectively. Both bodies start moving at the same time. After descending from the inclined plane, the bodies continue to move on a horizontal surface. Determine the velocity of these bodies after the collision, assuming the collision is completely elastic. [20 points]

Neglect friction forces. The acceleration due to gravity is g=9.81 m/s2g = 9.81 , \text{m/s}^2g=9.81m/s2.

Solution

To solve this problem, we will approach it in steps. Since the collision is completely elastic, we will use the laws of conservation of energy and momentum. The steps are as follows:


Step 1: Velocity of each body before the collision

Using the conservation of mechanical energy, the potential energy of each body at the top of the incline is converted into kinetic energy just before the collision.

For body m1=1kgm_1 = 1 \, \text{kg}:

m1gh1=12m1v12m_1 g h_1 = \frac{1}{2} m_1 v_1^2 19.812=121v121 \cdot 9.81 \cdot 2 = \frac{1}{2} \cdot 1 \cdot v_1^2 v12=29.812=39.24v1=39.246.26m/sv_1^2 = 2 \cdot 9.81 \cdot 2 = 39.24 \quad \Rightarrow \quad v_1 = \sqrt{39.24} \approx 6.26 \, \text{m/s}

For body mz=4kgm_z = 4 \, \text{kg}:

mzghz=12mzvz2m_z g h_z = \frac{1}{2} m_z v_z^2 49.811=124vz24 \cdot 9.81 \cdot 1 = \frac{1}{2} \cdot 4 \cdot v_z^2 vz2=29.811=19.62vz=19.624.43m/sv_z^2 = 2 \cdot 9.81 \cdot 1 = 19.62 \quad \Rightarrow \quad v_z = \sqrt{19.62} \approx 4.43 \, \text{m/s}

So, just before the collision: v1=6.26m/s,vz=4.43m/sv_1 = 6.26 \, \text{m/s}, \quad v_z = 4.43 \, \text{m/s}


Step 2: Velocities after the elastic collision

For a completely elastic collision, both momentum and kinetic energy are conserved. The final velocities v1v_1' and vzv_z' can be determined using the following formulas:

v1=(m1mz)v1+2mzvzm1+mzv_1' = \frac{(m_1 - m_z)v_1 + 2 m_z v_z}{m_1 + m_z} vz=(mzm1)vz+2m1v1m1+mzv_z' = \frac{(m_z - m_1)v_z + 2 m_1 v_1}{m_1 + m_z}

Substituting the values:

v1=(14)(6.26)+2(4)(4.43)1+4v_1' = \frac{(1 - 4)(6.26) + 2(4)(4.43)}{1 + 4} v1=36.26+244.435v_1' = \frac{-3 \cdot 6.26 + 2 \cdot 4 \cdot 4.43}{5} v1=18.78+35.445=16.6653.33m/sv_1' = \frac{-18.78 + 35.44}{5} = \frac{16.66}{5} \approx 3.33 \, \text{m/s}

vz=(41)(4.43)+2(1)(6.26)1+4v_z' = \frac{(4 - 1)(4.43) + 2(1)(6.26)}{1 + 4} vz=34.43+26.265v_z' = \frac{3 \cdot 4.43 + 2 \cdot 6.26}{5} vz=13.29+12.525=25.8155.16m/sv_z' = \frac{13.29 + 12.52}{5} = \frac{25.81}{5} \approx 5.16 \, \text{m/s}


Final Velocities

After the collision: v13.33m/s,vz5.16m/sv_1' \approx 3.33 \, \text{m/s}, \quad v_z' \approx 5.16 \, \text{m/s}


Would you like a detailed breakdown of any step or help with related concepts? Here are some related questions for further exploration:

  1. How would the velocities change if the collision were perfectly inelastic?
  2. What happens if friction on the incline is not negligible?
  3. How does the angle of the incline affect the velocities before the collision?
  4. What if the bodies were initially at different heights, e.g., h1=3mh_1 = 3 \, \text{m} and hz=2mh_z = 2 \, \text{m}?
  5. How would rotational motion influence the outcome, assuming the objects are spheres?

Tip: When solving elastic collision problems, always start by checking for energy and momentum conservation laws, as they are your key tools!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Conservation of Energy
Elastic Collision
Momentum
Kinematics

Formulas

mgh = 1/2 mv^2 (Conservation of Mechanical Energy)
v1' = ((m1 - m2) v1 + 2 m2 v2) / (m1 + m2) (Elastic Collision Formula)
v2' = ((m2 - m1) v2 + 2 m1 v1) / (m1 + m2) (Elastic Collision Formula)

Theorems

Conservation of Momentum
Conservation of Kinetic Energy

Suitable Grade Level

Grades 10-12