A golf ball of mass 50.0 grams is at rest on a tee. It is struck by a golf club with an average force of 850 Newtons. The ball accelerates to a speed of 68.0 m/sec. How long was the club in contact with the ball?

To find the time the club was in contact with the ball, we can use Newton's second law of motion, which relates force, mass, and acceleration. The formula is:

force = mass × acceleration

We are given the force, mass, and acceleration. We can rearrange the formula to solve for acceleration:

acceleration = force / mass

Substituting the given values:

acceleration = 850 N / 50.0 g

Note that we need to convert the mass from grams to kilograms, as the SI unit of force is Newtons and the SI unit of mass is kilograms.

mass = 50.0 g = 50.0 g × (1 kg / 1000 g) = 0.0500 kg

So, the acceleration is:

acceleration = 850 N / 0.0500 kg

Next, we can use the equation of motion to calculate the time taken:

final velocity = initial velocity + (acceleration × time)

The initial velocity is zero since the ball is at rest. The final velocity is given as 68.0 m/s. We can rearrange the equation to solve for time:

time = (final velocity - initial velocity) / acceleration

Plugging in the values:

time = (68.0 m/s - 0 m/s) / (850 N / 0.0500 kg)

Simplifying,

time = (68.0 m/s) × (0.0500 kg / 850 N)

Evaluating,

time = 0.399 seconds

Therefore, the club was in contact with the ball for approximately 0.399 seconds.

To calculate the time the club was in contact with the ball, we can use the equations of motion.

First, let's calculate the acceleration of the ball using Newton's second law of motion:

F = ma

Where F is the force applied, m is the mass of the ball, and a is the acceleration. Rearranging the equation, we have:

a = F / m

Substituting the given values, we have:

a = 850 N / 0.05 kg

a = 17,000 m/s²

Next, we can use the kinematic equation to find the time:

v = u + at

Where v is the final velocity, u is the initial velocity (which is zero since the ball was at rest), a is the acceleration, and t is the time. Rearranging the equation, we have:

t = (v - u) / a

Substituting the given values:

t = (68.0 m/s - 0) / 17,000 m/s²

t = 0.004 seconds

Therefore, the club was in contact with the ball for 0.004 seconds.

F = M*a.

a = F/M = 850/0.05kg = 17,000 m/s^2.

V = Vo + a*t = 68 m/s.
0 + 17,000t = 68,
t = 0.004 s.