Question

The head of a golf club strikes a 46G golf ball at rest. If the collision lasts 0.5 s and the ball is given a speed of 220 km/s, estimate the average force on the ball.
(A) 5270N
(B) 5520N
(C) 5320N
(D) 5620N

Answer 1

Hint: We will calculate the average force on the ball, so for this first we have to calculate the acceleration of the ball as it has a collision time of 0.5 s. We will calculate the acceleration of the ball using the kinematics formula \[v = u + at\] .
After finding the acceleration of the ball we put that value into the force formula, to calculate the average force on the ball.
And we will calculate the force by using \[F = ma\] , where we put the calculated acceleration and converted mass.

Complete step by step answer
We have to calculate the average force on the ball. So, for this we use the formula of force i.e. \[F = ma\] , where a is the acceleration of the ball and m is the mass of the ball.
Now the mass of the ball is given in the question, so we have to find the acceleration. For this we use the formula of motion when acceleration is constant i.e. \[v = u + at\] , where u is the velocity at time 0 , \[{\text{v}}\] is the velocity at the time \[{\text{t}}\]and a is the constant acceleration.
Now \[u = 0\] m/s , at \[t = 0\] S
 \[v = 220\] km/hr, we convert \[{\text{km/hr}}\] into m/s.
So, \[v = 61.1\] m/s at \[t = 0.5\] S
We put the above values into the equation \[v = u + at\] and find the acceleration
 \[v = u + at\]
 \[61.1 = 0.5a\]
After calculating we get \[a = 122.2\] m/s
We put this value of acceleration into the force formula to find the force on the ball.
 \[F = ma\]
 \[m = 46\] g, \[a = 122.2\dfrac{m}{{{s^{2}}}}\]
 \[F = 46 \times 122.2\] , as we put the value of acceleration we found and the value of mass is given.
So, the average amount of force acting on the ball is 5620 N.

So, the correct option is D.

Note: Always remember that the motion formula as they are very useful in solving the problems of motion in a straight line with constant acceleration and also only applicable when acceleration is constant.
The quantities u, v and a may take positive or negative values depending on whether they are directed along the positive or negative direction.
Also for freely falling bodies in the vertically downward direction we will denote acceleration as g which is equal to the value of \[9.8\dfrac{m}{{{s^2}}}\] and if we take a vertically upward direction acceleration will be \[a = - g\] .