Question

A car of mass m moving at a speed v1 collides and couples with the back of a truck of mass 3m moving initially in the same direction as the car at a lower speed v2. (Use any variable or symbol stated above as necessary.)

(b) What is the change in kinetic energy of the car–truck system in the collision?

Answer 1

initial momentum = m V1 + 3m V2
final momentum = 4 m V3
same final momentum as initial
so
V3 = (V1+3 V2)/4

initial Ke = .5 m( V1^2 + 3 V2^2)

final Ke = .5 (4m) (V3^2) = 2 m(V3^2) = 4 (.5m)V3^2

so change = .5m (V1^2+3V2^2 - 4 V3^2)
but
V3^2 = (V1^2 + 6V1V2 + 9V2^2)/16
4 V3^2 = (V1^2 + 6V1V2 + 9V2^2)/4
so change =
(1/2) m( (3/4)V1^2 -6/4 V1V2 +(3/4)V2^2)

= (m/8) (3 V1^2 - 6 V1V2 + 3 V2^2)

= (3 m/8)(V1-V2)^2

Answer 2

CHECK MY ARITHMETIC !!!
It came out awful coincidental though.

Answer 3

i cant seem to find the error but i still cant get the correct answer

Answer 4

Do it all over very carefully is all I can suggest.

Answer 5

You forgot the negative at the beginning in part B. Everything else is correct

Answer 6

assignment