Answer:
m_1 / m_2 = sqrt (1 / 2)
Explanation:
Given:
- Initial velocity of both skaters V_i = 0
- Velocity of skater 1 after push = V_1
- Velocity of skater after push = V_2
- Distance traveled by skater 1 = s_1
- Distance traveled by skater 2 = s_2
- s_1 = 2*s_2
- Accelerations of both skaters to halt is equal
Find:
What is the ratio m1/m2 of their masses
Solution:
- Apply conservation of momentum for two skaters just before and after the push as follows:
P_i = P_f
0 = m_1*V_1 - m_2*V_2
- Evaluate: m_1 / m_2 = ( V_2 / V_1 )
- Apply Conservation of Energy on both skaters as follows:
- Skater 1:
0.5*m_1*V_1^2 = u_k*m_1*g*s_1
-Simplify: 0.5*V_1^2 = u_k*g*(2*s_2)
- Skater 2:
0.5*m_2*V_2^2 = u_k*m_2*g*s_2
-Simplify: 0.5*V_2^2 = u_k*g*s_2
- Divide the two energy equations for skaters:
(V_1 / V_2)^2 = 2
(V_2 / V_1)^2 = 1 / 2
- simplify: (V_2 / V_1) = sqrt (1 / 2)
-Hence from earlier momentum conservation results:
m_1 / m_2 = ( V_2 / V_1 ) = sqrt (1 / 2)
