# How to Calculate Speed of Head of Driver After Collision?

• susan_khan
In summary: I'm still not clear on what this equation is supposed to do. I tried rereading the...problem but I'm still not clear on what this equation is supposed to do.
susan_khan
Homework Statement
A 0.045 kg golf ball is hit by a driver. The head of the driver has a mass of 0.15 kg, and travels at a speed of 56 m/s before the collision. The ball has a speed of 67 m/s as it leaves the clubface. What is the speed of the head of the driver immediately after the collision?
Relevant Equations
m1v1+m2v2 = m1v1f + m2v2f
vi1 +v1f = V2i +v2f
A 0.045 kg golf ball is hit by a driver. The head of the driver has a mass of 0.15 kg, and travels at a speed of 56 m/s before the collision. The ball has a speed of 67 m/s as it leaves the clubface. What is the speed of the head of the driver immediately after the collision?

I'm not sure where I'm going wrong but I first started off with this equation: m1v1+m2v2 = m1v1f + m2v2f. I was able to solve for v1f by using this equation and substituting it back into the main problem. (vi1 +v1f = V2i +v2f which equals 11 + v1f = v2f). After substuting this equation into the main equation I got v1f as 50.1 m/s and after subbing this number into this equation 11 + v1f = v2f I got v2f as 61.1 m/s. However the textbook says the answer is 36 m/s. What should I do?

susan_khan said:
Homework Statement:: A 0.045 kg golf ball is hit by a driver. The head of the driver has a mass of 0.15 kg, and travels at a speed of 56 m/s before the collision. The ball has a speed of 67 m/s as it leaves the clubface. What is the speed of the head of the driver immediately after the collision?
Relevant Equations:: m1v1+m2v2 = m1v1f + m2v2f
vi1 +v1f = V2i +v2f

A 0.045 kg golf ball is hit by a driver. The head of the driver has a mass of 0.15 kg, and travels at a speed of 56 m/s before the collision. The ball has a speed of 67 m/s as it leaves the clubface. What is the speed of the head of the driver immediately after the collision?

I'm not sure where I'm going wrong but I first started off with this equation: m1v1+m2v2 = m1v1f + m2v2f. I was able to solve for v1f by using this equation and substituting it back into the main problem. (vi1 +v1f = V2i +v2f which equals 11 + v1f = v2f). After substuting this equation into the main equation I got v1f as 50.1 m/s and after subbing this number into this equation 11 + v1f = v2f I got v2f as 61.1 m/s. However the textbook says the answer is 36 m/s. What should I do?
Use
m1v1+m2v2 = m1v1f + m2v2f
without dropping the masses. That's what you should do.

kuruman said:
Use
m1v1+m2v2 = m1v1f + m2v2f
without dropping the masses. That's what you should do.
Thank you for replying, but when solving I didn't drop the masses I just substituted 11 + v1f = v2f into this equation m1v1+m2v2 = m1v1f + m2v2f after inputting the values.

susan_khan said:
Thank you for replying, but when solving I didn't drop the masses I just substituted 11 + v1f = v2f into this equation m1v1+m2v2 = m1v1f + m2v2f after inputting the values.
Where does 11 + v1f = v2f come from? Why isn't v1f multiplied by m1 and v2f multiplied by m2? Please show your solution from its beginning, not just the middle part of it.

kuruman said:
Where does 11 + v1f = v2f come from? Why isn't v1f multiplied by m1 and v2f multiplied by m2? Please show your solution from its beginning, not just the middle part of it.
m1v1+m2v2 = m1v1f + m2v2f
(0.045)(67)+(0.15)(56) = 0.045v1f + 0.15v2f
3.015 + 8.4 = 0.045v1f + 0.15v2f
Before continuing I used this equation to solve for v1f
vi1 +v1f = V2i +v2f
67 + v1f = 56 +v2f
67 - 56 + v1f = 56 - 56 +v2f
11 + v1f = v2f
Sub this equation in:
3.015 + 8.4 = 0.045v1f + 0.15v2f
11.415 = 0.045v1f + 0.15(11 + v1f)
11.415 = 0.045v1f + 1.65 + 0.15v1f
11.415 -1.65 = 0.045v1f + 0.15v1f
9.765/0.195 = 0.195v1f/0.195
v1f = 50.1 m/s
Solve for v2f:
11 + 50.1 = v2f
v2f = 61.1 m/s

susan_khan said:
m1v1+m2v2 = m1v1f + m2v2f
(0.045)(67)+(0.15)(56) = 0.045v1f + 0.15v2f

Stop here.
At this point you figured out that you needed one more equation and you invoked this
vi1 +v1f = V2i +v2f
without explanation or justification. It is a bogus equation. The fact is that v1f is known. Read the problem carefully to see what it is. Put it in the momentum conservation equation and solve for v2f.

susan_khan said:
vi1 +v1f = V2i +v2f
This equation is only for the case of a perfectly elastic collision. The more general form, "Newton's Experimental Law", includes a coefficient of restitution.

Is golf a sport played by hitting balls in motion?

SammyS and nasu
kuruman said:
At this point you figured out that you needed one more equation and you invoked this
vi1 +v1f = V2i +v2f
without explanation or justification. It is a bogus equation. The fact is that v1f is known. Read the problem carefully to see what it is. Put it in the momentum conservation equation and solve for v2f.
I tried rereading the problem but I'm still unable to figure out the value for v1f. Could it be 0?

susan_khan said:
I tried rereading the problem but I'm still unable to figure out the value for v1f. Could it be 0?
There may be some confusion here because you do not state which object is 1 and which is 2 in your notation.
You are told the final speed of one and the initial speed of the other. The other initial speed should be obvious: this is golf, not tennis.

susan_khan said:
I tried rereading the problem but I'm still unable to figure out the value for v1f. Could it be 0?
Subscript "i" stands for "initial" and "f" stands for "final". How many objects are moving initially? How many objects are moving finally? Picture a golf ball being hit by a driver.

## 1. How far can a golf ball be hit by a driver?

The distance a golf ball can be hit by a driver varies depending on factors such as swing speed, launch angle, and wind conditions. On average, a professional golfer can hit a driver around 300 yards, while an amateur golfer may hit it around 200 yards.

## 2. What is the impact force of a golf ball being hit by a driver?

The impact force of a golf ball being hit by a driver can range from 2,000 to 4,000 pounds, depending on the speed of the swing and the angle of impact. This force is enough to cause damage to the ball and potentially cause it to deform or break.

## 3. How does the dimple pattern on a golf ball affect its flight when hit by a driver?

The dimple pattern on a golf ball helps reduce drag and increase lift, allowing the ball to travel farther and more accurately when hit by a driver. The dimples create a turbulent layer of air around the ball, reducing the wake behind it and allowing it to maintain a more stable trajectory.

## 4. Does the type of driver used affect the distance a golf ball can be hit?

Yes, the type of driver used can significantly affect the distance a golf ball can be hit. Factors such as the loft, weight distribution, and clubhead size can all impact the speed and trajectory of the ball, ultimately affecting the distance it travels.

## 5. How does the angle of the driver's face at impact affect the flight of the golf ball?

The angle of the driver's face at impact, also known as the loft angle, can greatly affect the flight of the golf ball. A higher loft angle will cause the ball to have a higher trajectory and more backspin, while a lower loft angle will result in a lower trajectory and less backspin. This can impact the distance and accuracy of the shot.

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