Math Problem Statement

A 1.2 kg object moving at 25.0 m/s in the x-direction collides and sticks to a 0.65 kg object moving at 35.0 m/s in the y-direction. Calculate the kinetic energy lost in the collision.

Solution

To calculate the kinetic energy lost in this collision, we need to follow these steps:

Step 1: Calculate the Initial Kinetic Energies

First, we calculate the initial kinetic energy of each object before the collision.

Object 1

  • Mass, m1=1.2kgm_1 = 1.2 \, \text{kg}
  • Velocity, v1x=25.0m/sv_{1x} = 25.0 \, \text{m/s}

The kinetic energy for object 1 is: KE1=12m1v1x2KE_1 = \frac{1}{2} m_1 v_{1x}^2 KE1=121.2(25.0)2KE_1 = \frac{1}{2} \cdot 1.2 \cdot (25.0)^2 KE1=0.6625=375JKE_1 = 0.6 \cdot 625 = 375 \, \text{J}

Object 2

  • Mass, m2=0.65kgm_2 = 0.65 \, \text{kg}
  • Velocity, v2y=35.0m/sv_{2y} = 35.0 \, \text{m/s}

The kinetic energy for object 2 is: KE2=12m2v2y2KE_2 = \frac{1}{2} m_2 v_{2y}^2 KE2=120.65(35.0)2KE_2 = \frac{1}{2} \cdot 0.65 \cdot (35.0)^2 KE2=0.3251225=398.125JKE_2 = 0.325 \cdot 1225 = 398.125 \, \text{J}

Total Initial Kinetic Energy

KEinitial=KE1+KE2KE_{\text{initial}} = KE_1 + KE_2 KEinitial=375+398.125=773.125JKE_{\text{initial}} = 375 + 398.125 = 773.125 \, \text{J}

Step 2: Calculate the Final Velocity of the Combined Object

Since the objects stick together, we have an inelastic collision. We use the conservation of momentum to find the final velocity of the combined object.

The combined mass MM of the two objects is: M=m1+m2=1.2+0.65=1.85kgM = m_1 + m_2 = 1.2 + 0.65 = 1.85 \, \text{kg}

The initial momentum in the xx- and yy-directions are: px=m1v1x=1.225.0=30.0kgm/sp_{x} = m_1 v_{1x} = 1.2 \cdot 25.0 = 30.0 \, \text{kg} \cdot \text{m/s} py=m2v2y=0.6535.0=22.75kgm/sp_{y} = m_2 v_{2y} = 0.65 \cdot 35.0 = 22.75 \, \text{kg} \cdot \text{m/s}

The final velocity components vfxv_{fx} and vfyv_{fy} of the combined object are: vfx=pxM=30.01.85=16.216m/sv_{fx} = \frac{p_{x}}{M} = \frac{30.0}{1.85} = 16.216 \, \text{m/s} vfy=pyM=22.751.85=12.297m/sv_{fy} = \frac{p_{y}}{M} = \frac{22.75}{1.85} = 12.297 \, \text{m/s}

Now we calculate the magnitude of the final velocity vfv_f: vf=vfx2+vfy2v_f = \sqrt{v_{fx}^2 + v_{fy}^2} vf=(16.216)2+(12.297)2v_f = \sqrt{(16.216)^2 + (12.297)^2} vf=262.97+151.24=414.21=20.36m/sv_f = \sqrt{262.97 + 151.24} = \sqrt{414.21} = 20.36 \, \text{m/s}

Step 3: Calculate the Final Kinetic Energy of the Combined Object

The final kinetic energy KEfinalKE_{\text{final}} of the combined object is: KEfinal=12Mvf2KE_{\text{final}} = \frac{1}{2} M v_f^2 KEfinal=121.85(20.36)2KE_{\text{final}} = \frac{1}{2} \cdot 1.85 \cdot (20.36)^2 KEfinal=0.925414.21=383.14JKE_{\text{final}} = 0.925 \cdot 414.21 = 383.14 \, \text{J}

Step 4: Calculate the Kinetic Energy Lost

The kinetic energy lost in the collision is the difference between the initial and final kinetic energies: ΔKE=KEinitialKEfinal\Delta KE = KE_{\text{initial}} - KE_{\text{final}} ΔKE=773.125383.14=389.985J\Delta KE = 773.125 - 383.14 = 389.985 \, \text{J}

Answer

The kinetic energy lost in the collision is approximately: ΔKE390J\Delta KE \approx 390 \, \text{J}

Would you like further details or have any questions?

Here are some related questions for practice:

  1. How would the kinetic energy lost change if the masses of the objects were different?
  2. What would happen to the kinetic energy lost if the objects did not stick together?
  3. How does the direction of the initial velocities affect the final kinetic energy?
  4. Can you apply the same approach to a collision where objects do not stick together?
  5. How does conservation of momentum differ in elastic and inelastic collisions?

Tip: In inelastic collisions, the kinetic energy is not conserved, but momentum always is.

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Math Problem Analysis

Mathematical Concepts

Physics
Kinetic Energy
Conservation of Momentum
Inelastic Collisions

Formulas

Kinetic Energy: KE = (1/2)mv^2
Momentum Conservation: p_initial = p_final
Velocity Magnitude: v_f = sqrt(v_fx^2 + v_fy^2)

Theorems

Conservation of Momentum

Suitable Grade Level

Grades 11-12