Please Help -- Golf Ball Hit by a Driver

  • #1
susan_khan
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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?
 

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  • #2
kuruman
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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.
 
  • #3
susan_khan
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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.
 
  • #4
kuruman
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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.
 
  • #5
susan_khan
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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
 
  • #6
kuruman
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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.
 
  • #7
haruspex
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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.
 
  • #8
Orodruin
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Is golf a sport played by hitting balls in motion?
 
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  • #9
susan_khan
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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?
 
  • #10
haruspex
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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.
 
  • #11
kuruman
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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.
 

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