on an ice skating rink, a girl of mass 50 kg stands stationary, face to face with a boy of mass 80 kg. the…

on an ice skating rink, a girl of mass 50 kg stands stationary, face to face with a boy of mass 80 kg. the children push off of one another, and the boy moves away with a velocity of +3 m/s. what is the final velocity of the girl? -1.9 m/s +1.9 m/s -4.8 m/s +4.8 m/s

on an ice skating rink, a girl of mass 50 kg stands stationary, face to face with a boy of mass 80 kg. the children push off of one another, and the boy moves away with a velocity of +3 m/s. what is the final velocity of the girl? -1.9 m/s +1.9 m/s -4.8 m/s +4.8 m/s

Answer

Explanation:

Step1: Apply conservation of momentum

The initial momentum of the system is 0 since they are stationary ($p_i = 0$). Let the mass of the girl be $m_1 = 50$ kg, her final velocity be $v_1$, the mass of the boy be $m_2=80$ kg and his final velocity be $v_2 = 3$ m/s. According to the law of conservation of momentum $p_i=p_f$, so $0=m_1v_1 + m_2v_2$.

Step2: Solve for the girl's velocity

We can re - arrange the equation $0=m_1v_1 + m_2v_2$ to solve for $v_1$. We get $v_1=-\frac{m_2v_2}{m_1}$. Substitute $m_1 = 50$ kg, $m_2 = 80$ kg and $v_2=3$ m/s into the formula: $v_1=-\frac{80\times3}{50}=-\frac{240}{50}=- 4.8$ m/s.

Answer:

-4.8 m/s