Answer:
0.05312 m
Explanation:
Given:
Length of platform L = 1.75 m
mass of ball m_b = 5.76 kg
mass of (people + platform) m_p = 184 kg
Initial Velocity of ball V_i,b = 0
Initial Velocity of ball (people + platform) V_i,p = 0
Find:
How far does the platform recoils to rest
Solution:
Using the conservation of momentum on ust before and after the ball was thrown P_i = P_f :
Where, P_i = 0 (initially at rest)
P_f = m_p*V_f,p + m_b * V_f,b
0 = m_p*V_f,p + m_b * V_f,b
V_f,p = - (m_b /m_p) * V_f,b
V_f,p = - (5.76 / 184)*V_f,b
V_f,p = - 0.0313*V_f,b ....1
The time the ball is in air:
t = L / (V_f,b - V_f,p) ...2
The distance that the platform moves d:
d = V_f,p *t ....3
Substitute 2 into 3
d = V_f,p*L /(V_f,b - V_f,p) .... 4
Solve 1 and 4 simultaneously :
d = - m_b*L / (m_b + m_p)
d = - 5.76*1.75 / (5.76 + 184)
d = -0.05312 m
The platform moves 0.05312 m to the opposite to which the ball is thrown.
