A 1200 kg car moving at 2.9 m/s is struck in the rear by a 2600 kg truck moving

Question

A 1200 kg car moving at 2.9 m/s is struck in the rear by a 2600 kg truck moving at 5.8 m/s.
(a) If the vehicles stick together after the collision, is the final kinetic energy of the car and truck together greater than, less than, or equal to the sum of the initial kinetic energies of the car and truck separately?

greater less equal

(b) Verify your answer to part (a) by calculating the initial and final kinetic energies of the system. To calculate the final kinetic energy you will need to find the final velocity of the vehicles after the collision using conservation of momentum.
Ki = kJ
Kf = kJ

Explanation / Answer

less (lost due to heat of combination) A 1200 kg car moving at 3.0 m/s is struck in the rear by a 2600 kg truck moving at 6.6 m/s. KI=1/2*m*v^2 1/2*1200*3^2+1/2*2600*6.6^2=62028 So to get Kf we use momentum mv=mv 1200*3+2600*6.6=(2600+1200)*v v=5.463.... KF=1/2*(2600+1200)*5.463^2=56704

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