Explanation:
It is given that,
Mass of golf club, m₁ = 210 g = 0.21 kg
Initial velocity of golf club, u₁ = 56 m/s
Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg
After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.
Initial momentum of golf ball, 
After the collision, final momentum 
Using the conservation of momentum as :
v = 63.91 m/s
So, the speed of the golf ball just after impact is 63.91 m/s. Hence, this is the required solution.
That's one of the three changes that are called "acceleration".
The other two are:
-- increase in the magnitude
-- change in direction.
Some might call a decrease in the magnitude "deceleration".
Answer
a) Using dimensional analysis we cannot derive the relation, But we can check the correctness of the formula.
now, L H S
s = distance
dimension of distance = [M⁰L¹T⁰]
now, equation on the right hand side
R H S
u = speed
u = m/s
Dimension of speed = [M⁰L¹T⁻¹]
dimension of time
t = sec
Dimension of time = [M⁰L⁰T¹]
Dimension of 'ut' = [M⁰L¹T⁻¹][M⁰L⁰T¹]
= [M⁰L¹T⁰]
now, acceleration= a
a = m /s²
dimension of acceleration = [M⁰L¹T⁻²]
dimension of (at²) = [M⁰L¹T⁻²][M⁰L⁰T¹][M⁰L⁰T¹]
= [M⁰L¹T⁰]
hence, the dimension are balanced.
so, L H S = R H S
b) Moment of inertia of hollow sphere = 
Moment of inertia of solid sphere = 
we know,
Torque is the force that causes rotation
If the same amount of torque is applied to both spheres the sphere with bigger moment of inertia would have smaller angular velocity.
Thus the solid sphere would accelerate more.
B.---A. warm water B. thermocline C. cold water
