Minta bantuannya ya kak,.

Minta bantuannya ya kak,.

MA =6kg
vA = 20m/s
mB= 4kg
vB = -30m/s

mAvA+mBvB = mAvA'+mBvB'
6.20 + 4.-30 =6.vA'+4vB'
120 -120 = 6vA'+4vB'
6vA' + 4vB' = 0 ………….[1]

lenting sempuran e=1
1 = (vB'-vA')/vA-vB
vA-vB = vB'-vA'
20–30 = vB'-vA'
50 =vB'-vA' ….. [2]
vB' = 50 + vA'

Eliminasi/SUbstitusi
6vA' + 4vB' = 0
6vA' + 4(50+vA') = 0
6vA' + 200 + 4vA' = 0
10vA' = -200
vA' = -20 m/s
vB' = 50+vA' = 50+-20 = 30m/s

Hukum kekekalan momentum

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

6×20 + 4×(-30) = 6v₁' + 4v₂'

6v₁' + 4v₂' = 0 …………..(1)

Tumbukan lenting sempurna (e=1)

$1 = frac{v'_{2} - v'_{1} }{20 - (-30)}$

-v₁' + v₂' = 50 …………….(2)

Eliminiasi persamaan (1) & (2)
6v₁' + 4v₂' = 0 (x1)
-v₁' + v₂' = 50 (x4)

  6v₁' + 4v₂'  = 0
- 4v₁' + 4v₂' = 200
______________ –
10v₁' = -200
  v₁' = -20 m/s ⇒ negatif artinya berlawanan arah
  v₂' = 50 – 20 = 30 m/s ⇒ positif artinya searah