A railroad engine, mass (M)= 60,000 kg, traveling at a speed of 25 m/s strikes a

Question

A railroad engine, mass (M)= 60,000 kg, traveling at a speed of 25 m/s strikes a car, mass (M)=6,000 kg, which is at initially at rest. After the collision, the railroad engine and the car stick together and move off down the tracks.

a. What were the initial momentum and the kinetic energy of the engine?


b.
i.) What was the momentum of the engine-car combination as they moved away from the collision?
ii.) What was their speed immediately after the collision?
iii.) What physical principle did you use to find these?


c.
i.) What was the kinetic energy of the engine-car combination immediately after the collision?
ii.) Was the kinetic energy conserved?
iii.) Was momentum conserved?

d. What impulse was delivered to the car by the engine? If the collision time was
t= 3* 10^-2 s, what was the average force exerted by the truck on the car?

Explanation / Answer

The mass of the engine is M = 60000 kg
The mass of the car is m = 6000 kg
The initial speed of the engine is ue = 25 m/s
The initial speed of the car is uc = 0 m/s

A)
The initial momentum of the engine is
Pe = Mue
Pe  = 60000*25
Pe  = 1500000 kg m/s

The initial kinetic energy of the engine

Ti = 0.5*M*ue2

Ti = 0.5*60000*625

Ti  = 18750000 J

B)According to the conservation of linear momentum,

Pi = Pf

The total momentum of the systen before collision is Pi = 1500000 kg m/s (car is at rest)

i)The momentum of the engine-car combination as they moved away from the collision is

Pf = 1500000 kg m/s

ii) Their speed immediately after the collision is given by

v = Pf /(M+m)

v = 1500000/66000

v = 22.73 m/s

iii)

The principle of conservation of linear momentum is used here, according to which "The total linear momentum of the system before collision is equalto that of after the collision"

C)

i) the kinetic energy of the engine-car combination immediately after the collision is

Tf = 0.5*(M+m)*v2

Tf = 0.5*66000*22.73^2

Tf = 17049545.7 J

ii)Clearly  Ti  > Tf

So the kinetic energy of the system is not conserved.

iii) The momentum of the system is conserved

D)

The time taken for the collision is t = 0.03 s

The impulse acted upon by the engine on the car will be

Fi = (Pf -Pi) for the car

Fi = m(v-0)

Fi = 6000*22.73

Fi = 136380 Ns

The average force on the car is

F = Fi /t

F = 136380/0.03

F = 4546000 N

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