Question

During an ice show a 70.0 kg skater leaps into the air and is caught by an initially stationary 55.0 kg skater. What is their final velocity in meters per second assuming negligible friction and that the 70.0-kg skater's original horizontal velocity was 4.00 m/s?

Answer 1

Answer:2.24 m/s

Explanation:

Let the mass of ice skater who is moving is $m_1=70kg$

mass of stationary skater, $m_2=55 kg$

conserving momentum

$m_1v+0=\left ( m_1+m_2\right )v_f$

$v_f=\frac{70\times 4}{125}$

$v_f$=$\frac{280}{125}=2.24 m/s$