A railroad car of mass 2.45 104 kg, is moving with a speed of 4.15 m/s.?

Question

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 an initial speed of 2.08 m/s.

Answer 1

Is the question the speed of the four cars together after collision?

m*4.15+3*m*2.08=4*m*v
solve for v
v=(4.15+3*2.08)/4

2.6 m/s

j

Answer 2

First. "Ithink Umno" starts off off by ability of adjusting your values with the aid of using a random assumption. How does that help you? 2d, he strikes directly to the "kinetic ability equation" and materials values in gadgets of Newtons. Kinetic ability (and all ability) is stated in Joules. Joules and Newtons, tension and ability are VERY distinctive, they do no longer equivalent one yet yet another, and you do no longer in many situations combination them in circumstances like those. that's no longer even the final place to start if it replaced into ultimate! the actual ideas --> (A) 2.575 m/s (B) ?KE = -40-one,057 J How and why. Given m1 = 2.58E+4 kg u1 = 4.12 m/s m2 = 3 * (2.58E+4 kg) = 77,4 hundred kg u2 = 2.06 m/s (A) submit-result velocity. in basic terms set up a typical Conservation of Momentum equation v * [m1 + m2] = m1u1 + m2u2 v * [ (2.58E+4 kg) + (77,4 hundred kg) ] = [ (2.58E+4 kg) * (4.12 m/s) ] + [ (77,4 hundred kg) * (2.06 m/s) ] v * [ 103,2 hundred kg ] = [ 106,296 kg-m/s ] + [ 159,444 kg-m/s ] v * [ 103,2 hundred kg ] = [ 265,740 kg-m/s ] v = [ 265,740 kg-m/s ] / [ 103,2 hundred kg ] v = 2.575 m/s (B) loss of kinetic ability unique KE is sum of first vehicle momentum plus momentum of different 3 vehicles m1 = 2.58E+4 kg u1 = 4.12 m/s KE1 = 0.5 * m * v^2 KE1 = 0.5 * (2.58E+4 kg) * (4.12 m/s)^2 KE1 = (12,900 kg) * (sixteen.ninety seven m^2/s^2) KE1 = 218,970 J m2 = 77,4 hundred kg u2 = 2.06 m/s KE2 = 0.5 * (77,4 hundred kg) * (2.06 m/s)^2 KE2 = (38,seven hundred kg) * (4.244 m^2/s^2) KE2 = 164,227 J Sum of finished preliminary ability KET = KE1 + KE2 KET = (218,970 J) + (164,227 J) KET = 383,197 J Now, the kinetic ability of the submit-result gadget m3 = m1 + m2 m3 = (2.58E+4 kg) + (77,4 hundred kg) m3 = 103,2 hundred kg v = 2.575 m/s (in basic terms have been given right here for the era of this) KE3 = 0.5 * (103,2 hundred kg) * (2.575 m/s)^2 KE3 = (51,600 kg) * (6.631 m^2/s^2) KE3 = 342,one hundred 40 J Kinetic ability loss is quite very final ability minus preliminary ability ?KE = KE3 - KET ?KE = (342,one hundred 40 J) - (383,197 J) ?KE = -40-one,057 J