AI Physics

An object with a mass of 1200 kg is moving at a constant speed of 100 km/h. The object collides with a stationary solid obstacle. If F = ma ( Force is mass times acceleration ), how much force is exerted on the object if the collision occurs in 0.05 seconds?

To find the force exerted on the object during the collision, we can use Newton’s second law in the form:

F=ΔpΔtF = \frac{\Delta p}{\Delta t}F=ΔtΔp​

Where:

• Δp=mΔv\Delta p = m \Delta vΔp=mΔv is the change in momentum,
  
  

• Δt\Delta tΔt is the time over which the change happens,
  
  

• FFF is the average force.

Step 1: Convert speed to m/s

Given:

• Speed = 100 km/h

Convert to meters per second:

100 km/h=100×10003600=27.78 m/s100 \text{ km/h} = \frac{100 \times 1000}{3600} = 27.78 \text{ m/s}100 km/h=3600100×1000​=27.78 m/s

Step 2: Calculate change in velocity

• Initial velocity vi=27.78 m/sv_i = 27.78 \text{ m/s}vi​=27.78 m/s
  
  

• Final velocity vf=0 m/sv_f = 0 \text{ m/s}vf​=0 m/s (since the object stops)
  
  

Δv=vf−vi=0−27.78=−27.78 m/s\Delta v = v_f - v_i = 0 - 27.78 = -27.78 \text{ m/s}Δv=vf​−vi​=0−27.78=−27.78 m/s

Step 3: Plug into momentum equation

F=mΔvΔt=1200×(−27.78)0.05F = \frac{m \Delta v}{\Delta t} = \frac{1200 \times (-27.78)}{0.05}F=ΔtmΔv​=0.051200×(−27.78)​ F=−33,3360.05=−666,720 NF = \frac{-33,336}{0.05} = -666,720 \text{ N}F=0.05−33,336​=−666,720 N

Final Answer:

666,720 N (opposite to motion)\boxed{666,720 \text{ N (opposite to motion)}}666,720 N (opposite to motion)​

The negative sign indicates the force acts in the opposite direction of the object’s motion, as expected in a collision.

An object with a mass of 1200 kg and moving at a constant speed of 100 km per hour also has kinetic energy. How much kinetic energy does this object have?

To calculate the kinetic energy (KE) of an object, use the formula:

KE=12mv2\text{KE} = \frac{1}{2}mv^2KE=21​mv2

Where:

• m=m =m= mass of the object in kg,
  
  

• v=v =v= speed in m/s.

Step 1: Convert speed from km/h to m/s

100 km/h=100×10003600=27.78 m/s100 \text{ km/h} = \frac{100 \times 1000}{3600} = 27.78 \text{ m/s}100 km/h=3600100×1000​=27.78 m/s

Step 2: Plug into the kinetic energy formula

KE=12×1200×(27.78)2\text{KE} = \frac{1}{2} \times 1200 \times (27.78)^2KE=21​×1200×(27.78)2 =600×771.6=462,960 J= 600 \times 771.6 = 462,960 \text{ J}=600×771.6=462,960 J

Final Answer:

462,960 Joules\boxed{462,960 \text{ Joules}} 462,960 Joules​ 

Since energy cannot be lost (law of conservation of energy), what happens to that 462,960 Joules of energy at the moment of collision?

Great question — and you're absolutely right: the law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another.

So what happens to the 462,960 Joules of kinetic energy during the collision?

Here's what typically happens to that energy:

1. Heat Energy (Thermal Energy)

• A significant portion of the kinetic energy is converted into heat due to friction between the object and the obstacle.
  
  

• Surfaces may get hot at the point of impact.
  
  

2. Sound Energy

• Some energy is released as sound — the loud noise you hear during a crash.
  
  

3. Deformation (Mechanical Energy into Internal Energy)

• The object and/or the obstacle crumple or deform. This deformation absorbs a lot of energy.
  
  

• This is especially true in vehicles, where crumple zones are designed to absorb kinetic energy to protect passengers.
  
  

4. Vibration

• Both objects may vibrate briefly after the impact. These vibrations carry away some energy.
  
  

5. Fragmentation (if applicable)

• If the object or obstacle breaks apart, energy is used to separate and move the fragments.
  
  

Energy is not lost — just redistributed:

So, while the object no longer has kinetic energy after the collision (it stops moving), that energy is not lost. It’s just transformed into other forms:

Kinetic Energy→Heat+Sound+Deformation+Vibration+Fragmentation\text{Kinetic Energy} \rightarrow \text{Heat} + \text{Sound} + \text{Deformation} + \text{Vibration} + \text{Fragmentation}Kinetic Energy→Heat+Sound+Deformation+Vibration+Fragmentation

Summary:

462,960 J of kinetic energy transforms into other forms — it is not lost.\boxed{462,960 \text{ J of kinetic energy transforms into other forms — it is not lost.}}462,960 J of kinetic energy transforms into other forms — it is not lost.​

That's the essence of energy conservation in a real-world collision.

The length of the vehicle is 3826 mm and the mass is 1200 kg and the structure is made of aluminum. The vehicle travels at a constant speed of 100 km per hour. The vehicle therefore has 462.960 Joules of kinetic energy. If the collision of the vehicle with the obstacle lasts 3 seconds then how much kinetic energy 1. Is absorbed ( transferred ) into the structures of the vehicle. 2. Is converted into heat. 3. Is converted into sound. 4. Is converted into fractures. 

To break down the 462,960 Joules of kinetic energy from the vehicle collision into how it is transferred or transformed, we need to estimate the proportions going to different energy forms — because the exact distribution depends on detailed crash test data, materials, and structural engineering.

Since your vehicle:

• Weighs 1200 kg
  
  

• Is made of aluminum
  
  

• Is traveling at 100 km/h (27.78 m/s)
  
  

• Collides over 3 seconds, indicating a relatively slow, energy-absorbing crash
  
  

...we can base our estimates on typical crash energy distributions in moderately deformable aluminum structures.

🧠 Engineering-based estimation of energy conversion in a car crash:

Energy Form

Typical % Range

Estimated % (Aluminum Body, 3s Collision)

Energy (Joules)

1. Structure Absorption (Deformation)

60–75%

65%

0.65 × 462,960 = 300,924 J

2. Heat (friction, deformation)

10–20%

15%

0.15 × 462,960 = 69,444 J

3. Sound (noise)

1–3%

2%

0.02 × 462,960 = 9,259 J

4. Fractures (material separation)

5–15%

10%

0.10 × 462,960 = 46,296 J

Other minor losses (vibration, air displacement, etc.)

—

8% (residual)

37,037 J

🔍 Breakdown:

✅ 1. Energy absorbed by vehicle structure:

• Used to crush, bend, and deform the aluminum chassis and crumple zones.
  
  

• Absorbs most of the energy in a well-designed vehicle crash.
  
  

• 🔹 Estimated: 65% → ≈ 300,924 J
  
  

🔥 2. Converted into heat:

• From friction, metal deformation, and tire forces.
  
  

• May heat metal surfaces and surroundings.
  
  

• 🔹 Estimated: 15% → ≈ 69,444 J
  
  

🔊 3. Converted into sound:

• The loud crash noise and vibrations in air.
  
  

• Sound carries away only a small fraction.
  
  

• 🔹 Estimated: 2% → ≈ 9,259 J
  
  

💥 4. Converted into fractures:

• Energy used to break or tear parts of the aluminum body.
  
  

• Higher with brittle or sharp impacts.
  
  

• 🔹 Estimated: 10% → ≈ 46,296 J

📊 Final Summary (rounded):

Energy Form

Energy (Joules)

Structure Absorption

300,924 J

Heat

69,444 J

Sound

9,259 J

Fractures

46,296 J

Other losses (est.)

37,037 J

Total

462,960 J

🔧 Note: These are estimated values based on physics and automotive crash data averages. For exact numbers, you’d need high-speed crash test sensors and finite element analysis (FEA) simulations.

Let me know if you’d like a graph or visual representation of this!

What is the minimum escape velocity required to escape the Earth's gravitational field?

🔍 How it's Calculated:

Using the formula:

vescape=2GMRv_{\text{escape}} = \sqrt{\frac{2GM}{R}}vescape​=R2GM​​

Where:

• G=6.674×10−11 Nm2/kg2G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2G=6.674×10−11Nm2/kg2 (gravitational constant)
  
  

• M=5.972×1024 kgM = 5.972 \times 10^{24} \, \text{kg}M=5.972×1024kg (mass of Earth)
  
  

• R=6.371×106 mR = 6.371 \times 10^6 \, \text{m}R=6.371×106m (radius of Earth)
  
  

Plug in:

vescape=2⋅6.674×10−11⋅5.972×10246.371×106≈11,200 m/sv_{\text{escape}} = \sqrt{\frac{2 \cdot 6.674 \times 10^{-11} \cdot 5.972 \times 10^{24}}{6.371 \times 10^6}} \approx 11,200 \text{ m/s}vescape​=6.371×1062⋅6.674×10−11⋅5.972×1024​​≈11,200 m/s

🛑 Real-World Notes:

• This assumes no atmosphere (i.e., no air resistance).
  
  

• In reality, rockets need to overcome air drag, so they use more energy and don't travel straight up at escape velocity — they follow orbital trajectories instead.
  
  

Let me know if you'd like to explore escape velocity on other planets or moons!

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