Question

Two trains of equal weight, moving with velocities of 30 miles an hour each and in opposite directions collide. Show that the loss of energy caused by the impact is the same as in the case of a train moving at 60 miles an hour striking another at rest. In the latter case, find the velocity with which the debris will be moved along the track. Also show that before impact the total energy in the one case is doubled that in the other

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Answer 1

considering the collision is perfectly in-elastic i.e., e=0;

case:1

speed of train A, V_A= 30 mph.

mass of train A be M_A = M

speed of train B, V_B= -30mph (-ve sign because as it is moving in opposite direction)

mass of train B be M_B= M

momentum is conserved,

M_AV_A + M_BV_B = (M_A+M_B)*V ('V' be the velocity of debris after collision)

30M-30M = (M_A+M_B)*V

V = 0;

Loss in Energy = Initial Energy - Final Energy

= 0.5*(M_AV_A² + M_BV_B²) - 0.5*(M_A+M_B)*V²

= 0.5*(2*M*30²)

L.E₁ = 900M ----------- 1

Intitial Energy (before collision) = I.E₁ = 0.5*2*M*30² = 900M ---------- 2

Final Energy (after collision) = F.E₁ = 0

case:2

speed of train A, V_A= 60 mph.

mass of train A be M_A = M

speed of train B, V_B= 0

mass of train B be M_B= M

momentum is conserved,

M_AV_A + M_BV_B = (M_A+M_B)*V ('V' be the velocity of debris after collision)

60M-0 = (M+M)*V

V = 30 mph ---------- 3

Loss in Energy = Initial Energy - Final Energy

= 0.5*(M_AV_A² + M_BV_B²) - 0.5*(M_A+M_B)*V²

= 0.5*(M*60²) - 0.5*(2*M*30²)

= 0.5*M*(3600-1800)

L.E₂ = 900M ------------ 4

Intitial Energy (before collision) = I.E₂ = 0.5*M*60² = 1800M ---------- 5

Final Energy (after collision) = F.E₂ =0.5*2*M*30² = 900M

from eq- 1 & 4, Loss of energy in both cases are equal, L.E₁ = L.E₂ = 900M;

from eq- 2 & 5, Initial Energy in case 2 is double the case 1, I.E₂ = 2*I.E₂;

from eq- 3, velocity of debris in case 2 is V = 30mph;