Charlotte, a 60.0-kg skin diver, shoots a 2.0-kg spear with a speed of 15 m/s at a fish who darts quickly away without getting hit. How fast does Charlotte move backwards when the spear is shot? (Hint: To start, Charlotte and the spear are together and both are at rest)

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

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.