The speed of the three couple car after the collision is 1.8 m/s and the total loss in kinetic energy of the system during the collision is 2.16 x 10⁴ J.
The mass M of the each railroad car is given to be 2 x 10³ kg and the speed U of the each railroad car is 3m/s. The speed V of the coupled car is given to be 1.20 m/s.
(a) According to the law Conservation of Linear Momentum, the linear momentum before and after the collision is always conserved.
So, we can write here,
MU + 2MV = (M+2M)V'
Where,
V' is the speed after collision.
U + 2V = 3V'
V' = (U+2V)/3
V' = (3+2.4)/3
V' = 1.8m/s.
(b) The loss in kinetic energy of the system can be calculated as,
∆K = final kinetic energy - initial kinetic energy
∆K = (1/2MU²+1/2×2MV²) - 1/2 × 3MV'²
Putting values,
∆K = (0.5×2.00×10⁴×(3.00)2+0.5×2(2.00×10⁴)(1.20)2)−0.5×3(2.00×10⁴)(1.8)2
∆K = 2.16 x 10⁴ J.
So, the loss in kinetic energy of the system after the collision is 2.16x10⁴ J.
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