A skater of mass 74.5 kg initially moves in a straight line at a speed of 5.10 m/s. The skater approaches a child of mass 37.5 kg, whom he lifts on his shoulders. Assuming there are no external horizontal forces, what is the skater's final velocity?
Linear momentum along the straight line is conserved, assuming neglible skate friction during the lift,
74.5*5.10 m/s = (74.5 + 37.5)* Vfinal
Solve for Vfinal.
Vfinal = 5.10 m/s(74.5/112)= 3.39 m/s
To find the skater's final velocity, we can use the principle of conservation of linear momentum. According to this principle, the total momentum before the child is lifted should be equal to the total momentum after the child is lifted.
The momentum (p) of an object can be calculated by multiplying its mass (m) by its velocity (v): p = m * v.
Before the child is lifted, the skater's momentum is given by: p_skater_initial = m_skater * v_skater_initial.
After the child is lifted, the skater's momentum is given by: p_skater_final = (m_skater + m_child) * v_skater_final.
Since there are no external horizontal forces, the momentum of the skater and the child before and after the child is lifted should be the same.
Therefore, we can set up an equation based on the principle of conservation of linear momentum:
m_skater * v_skater_initial = (m_skater + m_child) * v_skater_final.
Substituting the given values:
74.5 kg * 5.10 m/s = (74.5 kg + 37.5 kg) * v_skater_final.
Now, let's solve for v_skater_final:
v_skater_final = (74.5 kg * 5.10 m/s) / (74.5 kg + 37.5 kg).
v_skater_final = (379.95 kg * m/s) / 112 kg.
v_skater_final ≈ 3.39 m/s.
Therefore, the skater's final velocity after lifting the child on his shoulders is approximately 3.39 m/s.