The student 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

v2 = 27.3m/s

Explanation:

Assuming forward as positive.

Mass = m1 = 64kg

Let v be the common velocity of the student and the skateboard.

mass of skateboard = m2 = 5.94kg

v = 1.4m/s

Since the skateboard and the student are initially moving together at the same velocity their momentum together is

(m1 + m2) v

Let the final velocity of the student be v1 and the final velocity of the skateboard be v2

v1 = - 1.0m/s (falls backwards that's why the velocity is negative since we are assuming forward as positive)

Then from conservation of momentum, momentum before is equal to momentum after.

(m1 + m2) v = m1v1 + m2v2

m2v2 = (m1 + m2) v - m1v1

v2 = ((m1 + m2) v - m1v1) / m2

v2 = ((64 + 5.94) * 1.4 - 64 * (-1.0)) / 5.94

v2 = ((64 + 5.94) * 1.4 + 64*1.0) / 5.94

v2 = 27.3m/s