Math Problem Statement

A projectile of mass m = 0.2 kg and initial speed vi = 200 m/s hits a stationary target of mass M = 2 kg and gets imbedded in it. The change in kinetic energy ΔE during this collision is closest to

Solution

In this problem, we have an inelastic collision, where a projectile of mass m=0.2kgm = 0.2 \, \text{kg} hits a stationary target of mass M=2kgM = 2 \, \text{kg}, and they move together after the collision. The goal is to calculate the change in kinetic energy ΔE\Delta E during the collision.

Step 1: Initial kinetic energy of the projectile

The initial kinetic energy EiE_i of the projectile is given by:

Ei=12mvi2E_i = \frac{1}{2} m v_i^2

Substituting the given values:

Ei=12×0.2kg×(200m/s)2E_i = \frac{1}{2} \times 0.2 \, \text{kg} \times (200 \, \text{m/s})^2

Ei=0.1×40,000=4,000JE_i = 0.1 \times 40,000 = 4,000 \, \text{J}

Step 2: Final velocity after the collision

Since the projectile gets embedded in the target, the system (projectile + target) moves together after the collision. To find the final velocity vfv_f, we use the conservation of momentum:

mvi=(m+M)vfm v_i = (m + M) v_f

Solving for vfv_f:

vf=mvim+M=0.2×2000.2+2=402.218.18m/sv_f = \frac{m v_i}{m + M} = \frac{0.2 \times 200}{0.2 + 2} = \frac{40}{2.2} \approx 18.18 \, \text{m/s}

Step 3: Final kinetic energy of the system

The final kinetic energy EfE_f of the combined system (projectile + target) is:

Ef=12(m+M)vf2E_f = \frac{1}{2} (m + M) v_f^2

Substituting the values:

Ef=12×(0.2+2)×(18.18)2E_f = \frac{1}{2} \times (0.2 + 2) \times (18.18)^2

Ef=12×2.2×330.6363.66JE_f = \frac{1}{2} \times 2.2 \times 330.6 \approx 363.66 \, \text{J}

Step 4: Change in kinetic energy

The change in kinetic energy ΔE\Delta E is:

ΔE=EiEf=4,000363.663,636.34J\Delta E = E_i - E_f = 4,000 - 363.66 \approx 3,636.34 \, \text{J}

Conclusion:

The change in kinetic energy during the collision is approximately ΔE3,636J\Delta E \approx 3,636 \, \text{J}.


Would you like further details or have any questions on this? Here are some related questions for further understanding:

  1. How does the principle of conservation of momentum apply to inelastic collisions?
  2. What factors cause kinetic energy to decrease in inelastic collisions?
  3. Can you calculate the velocity of the target after the collision for different values of initial speed?
  4. How would the result change if the masses of the projectile and target were reversed?
  5. What is the difference in energy conservation between elastic and inelastic collisions?

Tip: In inelastic collisions, while momentum is always conserved, kinetic energy is not fully conserved due to energy transformation into other forms like heat and sound.

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Math Problem Analysis

Mathematical Concepts

Kinetic Energy
Conservation of Momentum
Inelastic Collision

Formulas

Initial kinetic energy: E_i = (1/2) m v_i^2
Final velocity: v_f = (m v_i) / (m + M)
Final kinetic energy: E_f = (1/2) (m + M) v_f^2
Change in kinetic energy: ΔE = E_i - E_f

Theorems

Conservation of Momentum
Energy Transformation in Inelastic Collisions

Suitable Grade Level

Grades 10-12