Two blocks of masses m1 = 2.00 kg and m2 = 3.70 kg are each released from rest at a height of h = 5.60 m on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision. (Let the positive direction point to the right. Indicate the direction with the sign of your answer.)

Question

(a) Determine the velocity of each block just before the collision.

(b) Determine the velocity of each block immediately after the collision.

(c) Determine the maximum heights to which m1 and m2 rise after the collision.

Answer 1

(a) The law of conservation of energy:
PE=KE
mgh =mv²/2
v=sqrt(2gh) = sqrt(2•9.8•5) = 9.9 m/s
m1==> v1 =+9.9 m/s
m2 ==> v2 = - 9.8 m/s
(b) the law of conservation of linear momentum:

After elastic collision
u1 = [(m1 - m2)/(m1 + m2)]•v1 + [ 2 •m2/(m1 + m2) ]•v2
u1 = [(2 kg – 3.7)/(2 + 3.7)]•(9.9) + [ 2•3.7 /(2 + 3.7)]•( - 9.9)=...
u2 = [2• m1 /(m1 + m2)]•v1 + [(m2 - m1) / (m1 + m2) ]•v2
u2= [2 •2 ) / (2 + 3.7)] •9.9 + [ (3.7- 2 ) / (2 + 3.7)]•( - 9.9 )=...
(c) law of conservation of energy:
KE=PE
mu²/2= mgh
h= u²/2g,
=>
h1= u1²/2g
h2= u2²/2g

Answer 2

mgh =mv²/2
v=sqrt(2gh) = sqrt(2•9.8•5.6) = 10.5 m/s
Correct further calculations...

Answer 3

why is it -9.9 when v2 is -9.8 m/s?