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.
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>
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 %
A 12100 kg railroad car is coasting on a level, frictionless track at a speed of...
A 12100 kg railroad car is coasting on a level, frictionless track at a speed of 19.0 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.
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