The boy soon loses his balance and falls backwards off the board at a velocity of 1.0 m/s. Assuming momentum is conserved in this process, what is the skateboard's new velocity in meters per second?

(a) W (s) = M*g

(b) 7.84kg

(c) - 10.07m/s

Explanation:

Parameters given:

Mass of boy, m = 79kg

Initial velocity of boy before jumping on the skateboard, u (b) = 1.55m/s

Final velocity of boy after jumping on the skateboard, v (b) = 1.41m/s

Mass of skateboard, M = ?

Initial velocity of skateboard = u (s)

Final velocity of skateboard = v (s)

(a) The weight of the skateboard is given as the mass of the skateboard multiplied by acceleration due to gravity. Mathematically,

W = M * g

Where g = acceleration due to gravity

(b) Using the law of conservation of momentum, we have that:

m*u (b) + M*u (s) = m*v (b) + m*v (s)

We know that after jumping on the skateboard, the boy and the skateboard have the same final velocity, hence,

v (b) = v (s) = v

=> m*u (b) + M*u (s) = (m + M) * v

Making M the subject of the formula, we have that:

M = m{u (b) - v} / v

M = 79 * (1.55 - 1.41) / 1.41

M = 7.84kg

(c) According to Newton's third law of motion, the momentum of the boy after falling must be equal to but opposite the momentum of the skateboard. This is the concept that birthed the law of conservation of momentum.

Momentum is given as:

P = m*v

Where P = momentum

m = mass of object

v = velocity of object

Therefore, applying Newton's third law of motion:

m * v (b) = - M * v (s)

Where

v (b) is now the velocity of the boy after failing off the skateboard = 1.0m/s

v (s) is now the velocity of the skateboard after the boy falls off.

Making v (s) subject of the formula, we have that:

v (s) = m * v (b) / M

v (s) = - (79 * 1.0) / 7.8

v (s) = - 10.07m/s

The negative sign means that that velocity of the boy is opposite the velocity of the skateboard.