A 1,185-kg car traveling initially with a speed of 25.0 m/s in an easterly direc
ID: 2191318 • Letter: A
Question
A 1,185-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 8,400-kg truck moving in the same direction at 20.0 m/s (see figure below). The velocity of the car right after the collision is 18.0 m/s to the east.
(a) What is the velocity of the truck right after the collision? m/s (east)
(b) How much mechanical energy is lost in the collision? (Use input values with an adequate number of significant figures to calculate this answer.) kJ Account for this loss in energy.
Explanation / Answer
A 1200kg car traveling initially with a speed of 25.0m/s in an easterly direction crashes into the rear end of a 9000kg truck moving in the same direction at 20.0m/s. the velocity of the car right after the collision is 18.0m/s to the east. What is the velocity of the truck right after the collision? how much mechanical energy is lost in the collision? Account for this loss in energy A 1200kg car traveling initially with a speed of 25.0m/s in an easterly direction crashes into the rear end of a 9000kg truck moving in the same direction at 20.0m/s. the velocity of the car right after the collision is 18.0m/s to the east. What is the velocity of the truck right after the collision? how much mechanical energy is lost in the collision? Account for this loss in energy A 1200kg car traveling initially with a speed of 25.0m/s in an easterly direction crashes into the rear end of a 9000kg truck moving in the same direction at 20.0m/s. the velocity of the car right after the collision is 18.0m/s to the east. What is the velocity of the truck right after the collision? how much mechanical energy is lost in the collision? Account for this loss in energy As the car has not deviated from the earlier direction, we assume that the collision is one-dimensional. It also makes calculations simple.As losses are present, we can take only conservation of momentum for our calculations.
M1 * U1 + M2 * U2 = M1 * V1 + M2 * V2.
1 stands for car, 2 for truck. M is mass, U is initial velocity (before collision) and V is final velocity (after collision).
M2 * V2 = 1200 * 25 + 9000 * 20 - 1200 * 18 = 188400
This gives V2 = 188400 / 9000 = 20.9333 m/s
To calculate losses, we will find the kinetic energies before & after collision. Any difference would give us the losses (in energy form).
(KE)1 = (KE)2 + Losses.
(KE)1 = 0.5 * (M1 * U1^2 + M2 * U2^2) = 2175000 J
(KE)2 = 0.5 * (M1 * V1^2 + M2 * V2^2) = 2166320 J
Losses = (KE)1 - (KE)2 = 8680 Joules
Hope this helped! As the car has not deviated from the earlier direction, we assume that the collision is one-dimensional. It also makes calculations simple.
As losses are present, we can take only conservation of momentum for our calculations.
M1 * U1 + M2 * U2 = M1 * V1 + M2 * V2.
1 stands for car, 2 for truck. M is mass, U is initial velocity (before collision) and V is final velocity (after collision).
M2 * V2 = 1200 * 25 + 9000 * 20 - 1200 * 18 = 188400
This gives V2 = 188400 / 9000 = 20.9333 m/s
To calculate losses, we will find the kinetic energies before & after collision. Any difference would give us the losses (in energy form).
(KE)1 = (KE)2 + Losses.
(KE)1 = 0.5 * (M1 * U1^2 + M2 * U2^2) = 2175000 J
(KE)2 = 0.5 * (M1 * V1^2 + M2 * V2^2) = 2166320 J
Losses = (KE)1 - (KE)2 = 8680 Joules
Hope this helped! As the car has not deviated from the earlier direction, we assume that the collision is one-dimensional. It also makes calculations simple.
As losses are present, we can take only conservation of momentum for our calculations.
M1 * U1 + M2 * U2 = M1 * V1 + M2 * V2.
1 stands for car, 2 for truck. M is mass, U is initial velocity (before collision) and V is final velocity (after collision).
M2 * V2 = 1200 * 25 + 9000 * 20 - 1200 * 18 = 188400
This gives V2 = 188400 / 9000 = 20.9333 m/s
To calculate losses, we will find the kinetic energies before & after collision. Any difference would give us the losses (in energy form).
(KE)1 = (KE)2 + Losses.
(KE)1 = 0.5 * (M1 * U1^2 + M2 * U2^2) = 2175000 J
(KE)2 = 0.5 * (M1 * V1^2 + M2 * V2^2) = 2166320 J
Losses = (KE)1 - (KE)2 = 8680 Joules
Hope this helped!
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.