Is my work correct?

Question

Two bodies hang on a string 2.1m long. The two bodies have a mass of 3.75kg and 2.4kg respectively. If the 3.75kg mass is raised to a height of 0.545m and released, and if upon impact the mass adhere.

Mass1 = 3.75 kg
Mass2 = 2.4 kg
String length = 2.1 m
M1 raised = 0.545m

a) Find the velocity of the 3.75kg mass before impact.
My ans.:Because the potential energy (Ep) of the 3.75kg mass will be equal to its kinetic energy (Ek), the following equation can be derived to find the velocity:
E_k=1/2 mv^2
E_p=mgh
1/2 mv^2=mgh
v^2=2gh
v=¡Ì2gh
v=¡Ì(2X9.81 m⁄s^2 X0.545m)
v=¡Ì(10.69 m^2⁄s^2 )
Velocity before impact=3.27 m⁄s (Ans.)

b) Find the common velocity of the mass after impact.
Lost inE_(k1 ) before impact=Gain inE_k2 after impact
1/2 M_1 (v_1 )^2=1/2 (M_1+M_2 )X(¡¼v_2)¡½^2
0.5X3.75kgX(3.27 m⁄s)^2=0.5X(3.75kg+2.4kg)X¡¼v_2¡½^2
20.049=3.075X¡¼v_2¡½^2
¡¼v_2¡½^2=2.049/3.075
¡Ì(¡¼v_2¡½^2 )=¡Ì6.52
Velocity after impact=2.55 m⁄s (Ans.)

c) Find the lost of kinetic energy on impact.
Loss of E_k1 on impact=¡¼Ek¡½_2 after impact
E_k2=1/2(m_1+m_2)¡¼v_2¡½^2
E_k2=0.5X(3.75kg+2.4kg)X(2.55 m⁄s)^2
E_k2=3.075X¡¼2.55¡½^2
Kinetic Energy after impact=19.99J (Ans.)

d) Find the height through which the center of gravity of the system will rise.
E_p1 before impact=E_p2 after impact
m_1 gh_1=(m_1+m_2 )gh_2
3.75kgX9.81X0.545m=(3.75kg+2.4kg)X9.81Xh_2
20.05=60.33Xh_2
h_2=20.05/60.33
Final height the system will rise=0.332m (Ans.)

e) Find the tension on the string of the 3.75kg mass before impact.
Tension on the string is equal to the centrifugal force exerted by the 3.75kg mass.
F_c=(mv^2)/r
F_c=(3.75X¡¼3.27¡½^2)/2.1
F_c=40.098/2.1
F_c=19.09N (Ans.)