Answer:
67.22 m/s
Explanation:
First, find deceleration which is equivalent to, [tex]a=\mu_k g[/tex] where [tex]\mu_k[/tex] is coefficient of kinetic friction and g is acceleration due to gravity. Taking g as 32.2 then a=0.75*32.2=24.15 ft/s2
From kinematics
[tex]v^{2}=u^{2}-2as[/tex] where v is final velocity and u is initial velocity. In this case, initial velocity is the common velocity and the final velocity is zero since they come to rest. S is the distance moved and a is deceleration. Substituting v with 0 and making u the subject then [tex]u=\sqrt{2as}[/tex] and by substitution [tex]u=\sqrt{2\times 24.15\times 17.5}\approx 29.073 ft/s[/tex]
Converting ft per s to miles per hour, we multiply the above by 0.681818 hence u=29.073*0.681818=19.8225 mph
From the law of conservation of linear momentum, p=mv where m is mass and v is velocity, the sum of initial and final momentum are equal. In this case
[tex]m_1v_1+m_2v_2=(m_1+m_2)u[/tex]
By substitution considering the u is already known as 19.8225 mph
[tex]1515v_1-(1125*44)=(1515+1125)*19.8225\\v_1=\frac {(1515+1125)*19.8225+(1125*44)}{1515}\approx 67.22\ mph[/tex]
