How Is the Average Velocity of 22 m/s Calculated in a Golf Ball Strike?

AI Thread Summary
The average velocity of a golf ball struck by a club is calculated as 22 m/s, derived from its initial velocity of 44 m/s during contact. The contact time is determined using the formula t = change in x / average velocity, where the change in x is the radius of the ball (0.05 m). This results in a contact time of approximately 0.00227 seconds. The momentum of the 50g golf ball after being struck is 2.2 kg m/s, indicating significant motion. Understanding these concepts is crucial for analyzing the dynamics of objects in motion.
mushhhy
Messages
7
Reaction score
0
A 50g golf ball is struck w a club. the ball leaves the face of the club w a velocity of 44m/s

Pf = mvf = (50 x 10^-3 kg) (44 m/s) = 2.2 kg m/s

From this I must find the contact time t = change in x /average velocity

The change in x in the radius of the ball .05m

The average velocity provided in the solution is 22 m/s

I do not understand how this figure of 22 m/s was obtained from the original velocity of 44 m/s

Help anyone
 
Physics news on Phys.org
Where did you get the change in x is .05 m? That information wasn't given to you in the problem... is
A 50g golf ball is struck w a club. the ball leaves the face of the club w a velocity of 44m/s
all you were given?
 


The 22 m/s is the average velocity because the ball is traveling at a constant speed of 44 m/s for a short period of time during the contact with the club. This means that the initial velocity and final velocity are both 44 m/s, making the average velocity 22 m/s.

To find the contact time, we use the formula t = change in x / average velocity. In this case, the change in x is the radius of the ball, which is 0.05m. So, t = 0.05m / 22 m/s = 0.00227 seconds. This means that the contact time between the ball and the club is only 0.00227 seconds.

Momentum is a measure of an object's motion and is calculated by multiplying its mass and velocity. In this case, the momentum of the golf ball after being struck by the club is 2.2 kg m/s. This shows that the ball has a significant amount of motion and it will continue to move at a constant speed unless acted upon by an external force.

Overall, the concept of momentum and contact time is important in understanding the motion of objects and how they interact with each other. In this example, we can see how a small and light object like a golf ball can have a large momentum and how the contact time between the ball and the club is very short.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top