Mechanical energy lost in collision?
a railroad car of mass 2.5 times 10^4 kg is moving with speed of 4.00 m/s. it collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with initial speed of 2 m/s.
How much mechanical energy is lost in the collision?
More Questions From This User See All
Answers & Comments
Verified answer
Total initial momentum = 2.5*4 + 3*2.5*2
= 25 e4 kg m/s
Let final velocity = v
Final momentum = 4*2.5*v = 10v
10v = 25 for momentum conservation
thus v = 2.5 m/s
Initial KE = 2.5/2 * 4^2 + 3*2.5/2 * 2^2
= 35e4 J
Final KE = 4*2.5/2 * 2.5^2 = 31.25e4 J
Energy loss = 3.75e4 J
= 37.5 kJ
1º)conservation of momentum
2.5*10^4*4 +3*2.5*10^4 *2 = 4*2.5*10^4*V
10+15= 10V so V = 2.5m/s
Initial kinetic energy =( 2.5/2 *16 +7.5/2*4*10^4=35*10^4 Joule
final kinetic energy = 10/2*6.25*10^4 = 31.25 *10^4 Joule
Energy lost = 3.75*10^4 Joule
M1= 2.5x10^4 kg
V1= +4
M2= 3(2.5x10^4) kg
V2= +2
M1v1 + M2v2= M(1+2)V*, V* being v prime, or the new velocity after totally inelastic collision. Plug in, and solve for V*.
Kinetic energy before= 1/2 (m1)v^2 + 1/2 (m2)v^2
Kinetic energy after= 1/2 (M1+m2) V*^2
K lost= (Kbefore - Kafter) j, or joules
yeah, what dr. d said