A railroad freight car of mass 3.06 *10^4 kg collides with a stationary caboose

Question

A railroad freight car of mass 3.06 *10^4 kg collides with a stationary caboose car. They couple together, and 17.0% of the initial kinetic energy is transferred to thermal energy, sound vibrations, and so on. Find the mass of the caboose.
____kg




Two 1.8 kg bodies, A and B, collide. The velocities before the collision are vA = (14i + 25j) m/s and vB = (-10i + 15.0j) m/s. After the collision, v'A = (-6.0i + 20j) m/s.

(a) What is the final velocity of B?
( 10 i +20 j) m/s (is this correct?)

(b) What is the change in the total kinetic energy (including sign)?
___ J



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Explanation / Answer

1st part

Let the initial speed = Vi & Final Velocity = Vf & mass of the caboose = m*104 kg

3.06*104*Vi = (3.06+m)*104 *Vf ( conservation of momentum) ------- eqn 1

Total kinetic energy left = 100-17 = 83%

3.06*104*Vi2*83/100 = (3.06+m)*104*Vf2

=> (Vi/Vf)2 = (3.06+m)*100/3.06*83 --------- eqn 2

from eqn 1

=> (Vi/Vf) = 3.06+m/3.06

Putting this in eqn 2

=> 3.06+m/3.06 = 100/83 => 83*m = 306 - 83*3.06 => m = 0.627

So Mass of Caboose = m*104 kg = 6.27*103 kg

2nd Part

Since masses are equal by conservation of momentum

(a) v'B = vA+vB-v'A = (14i+25j)+ (-10i+15j) - (-6i+ 20j) = (10 i+ 20j) m/s

(b) |vA| = 142+252 = 821 m/s

|vB| = 325 m/s

|v'A| = 436 m/s

|v'B| = 500 m/s

Initial K.E.= (1/2)*1.8*(821 + 325) = 1031.4 J

Final K.E. = (1/2)*1.8*(436+500) = 842.4 J

So Change = Final - Initial = - 189 J

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