Question

The head of a golf club moving at 45. 0 m/s strikes a golf ball (mass 46. 0 g) resting on a tee. The effective mass of the clubhead is 220 g. (a) with what speed does the ball leave the tee? (b) with what speed would it leave the tee if you doubled the mass of the clubhead? if you tripled it? what conclusions can you draw about the use of heavy clubs? assume that the collisions are perfectly elastic and that the golfer can bring the heavier clubs up to the same speed at impact. See question 13

Answer 1

We can use the conservation of momentum and energy to solve this problem. Since the collision is perfectly elastic, both momentum and kinetic energy are conserved.

Let's first find the initial velocity of the combined club and ball system:

m_clubhead = 220 g = 0.22 kg (effective mass of the clubhead)

m_ball = 46.0 g = 0.046 kg

v_clubhead = 45.0 m/s

The initial momentum of the system is:

p_i = m_clubhead * v_clubhead = 0.22 kg * 45.0 m/s = 9.90 kg·m/s

Since momentum is conserved, the final momentum of the system is also equal to 9.90 kg·m/s:

p_f = m_clubhead * v_clubhead' + m_ball * v_ball'

where v_clubhead' and v_ball' are the velocities of the clubhead and ball after the collision, respectively.

The initial kinetic energy of the system is:

KE_i = 1/2 * m_clubhead * v_clubhead^2 = 0.5 * 0.22 kg * (45.0 m/s)^2 = 222.75 J

Since kinetic energy is conserved, the final kinetic energy of the system is also equal to 222.75 J:

KE_f = 1/2 * m_clubhead * v_clubhead'^2 + 1/2 * m_ball * v_ball'^2

Now we can solve for v_ball':

p_f = m_clubhead * v_clubhead' + m_ball * v_ball'

9.90 kg·m/s = 0.22 kg * v_clubhead' + 0.046 kg * v_ball'

KE_f = 1/2 * m_clubhead * v_clubhead'^2 + 1/2 * m_ball * v_ball'^2

222.75 J = 0.5 * 0.22 kg * v_clubhead'^2 + 0.5 * 0.046 kg * v_ball'^2

We have two equations and two unknowns (v_clubhead' and v_ball'), so we can solve for v_ball':

v_ball' = (p_f - m_clubhead * v_clubhead') / m_ball

Substituting this expression into the energy conservation equation and solving for v_clubhead', we get:

v_clubhead' = sqrt(2/m_clubhead * (m_ball * v_ball'^2 + KE_i - KE_f))

Now we can use this equation to answer the questions:

(a) With the given data, we get:

v_ball' = (9.90 kg·m/s - 0.22 kg * 45.0 m/s) / 0.046 kg = 93.70 m/s

v_clubhead' = sqrt(2/0.22 * (0.046 kg * (93.70 m/s)^2 + 222.75 J - 222.75 J)) = 45.0 m/s

Therefore, the ball leaves the tee with a speed of 93.70 m/s.

(b) If we double the mass of the clubhead, the effective mass becomes 440 g = 0.44 kg, and the initial momentum of the system doubles to 19.80 kg·m/s. Using the same equations as before, we get:

v_ball' = (19.80 kg·m/s - 0.44 kg * 45.0 m/s) / 0.046 kg = 187.41 m

Step-by-step explanation: