Respuesta :
Answer:
Charlotte move backwards with 0.5 m/s when the spear is shot
Explanation:
Before collision;
m₁ = 60 kg
m₂ = 2 kg
v = 0 (since, Charlotte and the spear are together and both are at rest)
After collision
m₁ = 60 kg
m₂ = 2kg
v₁ = ?
v₂ = 15 m/s
From principle of conservation of linear moment, total momentum before collision must be equal to total momentum after collision.
v(m₁ + m₂) = m₁v₁ + m₂v₂
0 (60+2) = 60v₁ + 2 x 15
0 = 60v₁ + 30
60v₁ = -30
v₁ = - 30/60
v₁ = - 0.5 m/s
Therefore, Charlotte move backwards with 0.5 m/s when the spear is shot.
Answer:
0.5 m/s
Explanation:
From Newton's third law of motion,
Total momentum before collision = Total momentum after collision
mu+m'u' = mv+m'v'................. Equation 1
Where m = mass of the Charlotte, m' = mass of the spear, u = initial velocity of the Charlotte, u' = initial velocity of the spear, v = Final velocity of the Charlotte, v' = Final velocity of the spear.
Note: Since the Charlotte and the spear where initially at rest,
0 = mv+m'v'
-mv = m'v'
make v the subject of the equation
v = -(m'v'/m)..................... Equation 2
Given: m = 60 kg, m' = 2.0 kg, v' = 15 m/s
Substitute into equation 2
v = -(2×15/60)
v = -0.5 m/s.
The negative sign shows that the Charlotte moves backward, which is in opposite direction to the motion of the spear