A golf ball of mass 0.05 kg is struck by a golf club of mass m kg with a speed of 50m/s will have an initial speed of v(m)=83m/(m+0.05) m/s. show that v'(m)>0 and interpret this result in context with the application. Compare v'(0.15) and v'(0.20) in context of the application as well.

Question

A golf ball of mass 0.05 kg is struck by a golf club of mass m kg with a speed of 50m/s will have an initial speed of v(m)=83m/(m+0.05) m/s. show that v'(m)>0 and interpret this result in context with the application. Compare v'(0.15) and v'(0.20) in context of the application as well.

Answer 1

v(m) = 83m / (m+0.05)

dv/dm = 1660/(20x+1)^2 = 4.15/(x+0.05)^2

clearly v' > 0

v(m) = 83 - 83/(20m+1)

clearly the maximum possible initial velocity is 83m/s, which is achievable only with a golf club of infinite mass.

naturally, a heavier club hits the ball faster

Answer 2

To show that v'(m) > 0, we need to find the derivative of v(m) with respect to m, and then analyze its sign. Let's start by finding the derivative.

Given:
v(m) = 83m / (m + 0.05)

To find v'(m), we can use the quotient rule for differentiation:

v'(m) = [83(m + 0.05) - 83m] / (m + 0.05)^2

Now, simplify the expression:

v'(m) = [83m + 4.15 - 83m] / (m + 0.05)^2
= 4.15 / (m + 0.05)^2

Since the numerator (4.15) is a positive constant and the denominator (m + 0.05)^2 is always positive, v'(m) is positive for all values of m greater than zero. Therefore, we can conclude that v'(m) > 0.

Now, let's interpret this result in the context of the application. The positive value of v'(m) implies that as the mass of the golf club (m) increases, the initial speed of the golf ball (v(m)) also increases. This suggests that using a golf club with a greater mass would result in a faster initial speed for the golf ball.

Now, let's compare v'(0.15) and v'(0.20) in the context of the application:

v'(0.15) = 4.15 / (0.15 + 0.05)^2 = 4.15 / 0.20^2 = 103.75
v'(0.20) = 4.15 / (0.20 + 0.05)^2 = 4.15 / 0.25^2 = 66.4

When the mass of the golf club is 0.15 kg, the rate of change of the initial speed (v'(0.15)) is 103.75. On the other hand, when the mass of the golf club is 0.20 kg, the rate of change of the initial speed (v'(0.20)) is 66.4.

This comparison indicates that increasing the mass of the golf club from 0.15 kg to 0.20 kg leads to a lower rate of change in the initial speed of the golf ball. In other words, the effect of increasing the mass on the initial speed decreases as the mass of the golf club increases.