Answer:
The other student (59kg) moves right at 7.44 m/s
Explanation:
Given;
mass of the first student, m₁ = 77kg
mass of the second student, m₂ = 59kg
initial velocity of the first student, u₁ = 0
initial velocity of the second student, u₂ = 0
final velocity of the first student, v₁ = 5.7 m/s left
final velocity of the second student, v₂ = ? right
Apply the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(77 x 0) + (59 x 0) = (-77 x 5.7) + (59 x v₂)
0 = - 438.9 + 59v₂
59v₂ = 438.9
v₂ = 438.9 / 59
v₂ = 7.44 m/s to the right
Therefore, the other student (59kg) moves right at 7.44 m/s
