Question

Two skaters, both of mass 75 kg, are on skates on a frictionless ice pond. One skater throws a 0.3-kg ball at 5 m/s to his friend, who catches it and throws it back at 5 m/s. When the first skater has caught the returned ball, what is the velocity of each of the two skaters?

Answer 1

To solve this problem it is necessary to apply the concepts related to the momentum. The moment can be basically defined as the product of the mass and velocity of an object. Mathematically it is expressed as

$P = mv$

Where,

m = mass

V = Velocity

From the statement it is noted that there is a conservation of Momentum but given in different directions, to which we could affirm that there is the momentum of throw and momentum of return

$P_T = P_t+P_r$

Our values are given as

$m_S = 75kg$

$m_b = 0.3Kg$

$v = 5m/s$

Under the values given the moment with respect to the ball - and which is subsequently transmitted to people - it would be given by

$P_T = P_t+P_r$

$P_T = m_bv_t+m_bv_r$

$P_T = (0.3)(5)+(0.3)(5)$

$P_T = 3kg\cdot m/s$

If the moment is conserved the speed of each of the Skaters would be:

$P = mv$

$v = \frac{P}{m}$

$v = \frac{3}{75}$

$v = 0.04m/s$

Therefore the velocity of each of the two skaters is 0.04m/s moving apart.