Question

A 12100 kg railroad car is coasting on a level, frictionless track at a speed of 19 m/s when a 4790 kg load is dropped onto it.

If the load is initially at rest, find the new speed of the car and the % change of the kinetic energy.

Answer 1

using momentum conservation in horizontal direction,

12100 x 19 + 4790 x 0 = (12100 + 4790)v

v =13.61 m/s

change in K.E. = k.E. final - initial / initial x 100

= [(12100 + 4790)x13.61^2 /2] - [ 12100x 19^2 /2 ] x 100 /[ 12100x 19^2 /2 ]

= -28.38 %

minus sign indicates decreament in K.E>

Answer 2

m1v1=m2v2
v2=(m1/m2)v1
?K=1/2m2v2^2-1/2m1v1^2
?K=1/2(m1^2/m2)v1^2-1/2m1v1^2
?K=1/2m1v1^2(m1/m2-1)
?K=0,5*12100*19^2{[12100 / (12100+4790)]-1} = -619,396 J
?K%=(?K/Ki)*100
Ki=1/2m1v1^2=2,184,050 J
?K%=(-619,396/2,184,050)*100 = -28,4 %