Question

On an icy road, a 1100 kg car moving at 55 km/h strikes a 480 kg truck moving in the same direction at 37 km/h . The pair is soon hit from behind by a 1300 kg car speeding at 49 km/h. If all three vehicles stick together, what is the speed ofthe wreckage?

Answer 1

Answer:

Speed of the wreckage = 49.29 km/hr

Explanation:

This question is solved simply by using the conservation of momentum law.

The momentum of the three moving bodies are calculated below:

Momentum of Car 1 : 1100 * 55 = 60500 kg.km/hr

Momentum of Truck : 480 * 37 = 17760 kg.km/hr

Momentum of Car 2 : 1300 * 49 = 63700 kg.km/hr

Total mass of all three vehicles: 1100 + 480 + 1300 = 2880 kg

The final momentum equals the initial momentum if it is conserved. Thus we have the following equation:

Final Momentum = Initial Momentum

Final Velocity * Total mass = Momentum of all three vehicles combined

Final Velocity * 2880 = 60500 + 17760 + 63700

Final Velocity = 49.29 km/hr

Answer 2

Answer:

the speed of the wreckage is 48.39km/hr

Explanation:

Given that ,

Mass of first car, m₁ =1100kg

Speed of the first car, v₁ = 55km/hr

Mass of second car, m₂ = 480kg

Speed of second car, v₂ = 37km/hr

Mass of third car, m₃ =1300kg

Speed of third car, v₃ = 47km/hr

In the first case,

$m_1v_1+m_2v_2=(m_1+m_2)v$

$(1100\times 55) + (480\times 37)=(1100+480)v\\\\v =\frac{78260}{1580} \\\\v =49.53km/hr$

In the second case

$(m_1+m_2)v+m_3v_3=(m_1+m_2+m_3)v$

where v is the velocity of the combined system

$(1100+480)\times 49.53=(1300\times 47)=(1100+480+1300)\\\\(78257.4+61100)=2880\\\\v = \frac{139357.4}{2880} \\\\v =48.39km/hr$

Thus , the speed of the wreckage is 48.39km/hr