Solving a Physics Problem: Boy & Girl on a Frozen Pond

AI Thread Summary
A 55-kg girl throws a 1.8-kg ice ball at 7.5 m/s to a 75-kg boy on a frictionless pond. Using conservation of momentum, the girl's final velocity after throwing the ball is calculated to be 0.25 m/s, while the boy's final velocity upon catching the ball is 0.18 m/s. The discussion emphasizes the application of momentum conservation principles to solve the problem. Participants clarify the equations used and confirm the calculations. The problem is resolved simply, highlighting the effectiveness of momentum conservation in physics.
Mltn12
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Homework Statement



A 55-kg girl is standing near and to the left of a 75-kg boy on the frictionless surface of a frozen pond. The girl tosses a 1.8-kg ice ball to the girl with a horizontal speed of 7.5 m/s and he catches it. What are the velocities of the boy and the girl immediately after the boy catches the ball?

Homework Equations



MgVog+MbVb=0
Vfg=MballVb-Mb / Mg

The Attempt at a Solution



1.8kg x 7.5 m/s - 75kg / 55kg
= -1.1

Vfb=1/2 MbVb/ ( -1.1Mg-Mb)
= -10x10^-2

I don't think I did this correctly, can someone walk me through this?
 
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Try solving it for only the girl. If a 55 kg girl throws a 1.8kg ball at 7.5 m/s what will be her final speed? (When she starts at rest)
 
Mltn12 said:

Homework Statement



A 55-kg girl is standing near and to the left of a 75-kg boy on the frictionless surface of a frozen pond. The girl tosses a 1.8-kg ice ball to the girl with a horizontal speed of 7.5 m/s and he catches it. What are the velocities of the boy and the girl immediately after the boy catches the ball?


Homework Equations



MgVog+MbVb=0
Vfg=MballVb-Mb / Mg

The Attempt at a Solution



1.8kg x 7.5 m/s - 75kg / 55kg
= -1.1

Vfb=1/2 MbVb/ ( -1.1Mg-Mb)
= -10x10^-2

I don't think I did this correctly, can someone walk me through this?

Hello Mltn12. Welcome to PF !
 
V=Vo+at?
A=t/v, but I don't have t.
 
Nathanael said:
Try solving it for only the girl. If a 55 kg girl throws a 1.8kg ball at 7.5 m/s what will be her final speed? (When she starts at rest)

V=Vo+at?
A=t/v, but I don't have t.
 
Mltn12 said:

Homework Equations



MgVog+MbVb=0

What about this equation? That is conservation of momentum, right? Shouldn't it also apply to the girl and the ball?
 
Nathanael said:
What about this equation? That is conservation of momentum, right? Shouldn't it also apply to the girl and the ball?


Vfg=MballVb-Mb / Mg

1.8kg x 7.5 m/s - 75kg / 55kg
= -1.1
 
Mltn12 said:
Vfg=MballVb-Mb / Mg

1.8kg x 7.5 m/s - 75kg / 55kg
= -1.1

Where did you get V_{girl}=M_{ball}V_{ball}-\frac{M_{ball}}{M_{girl}}?

Use the other equation M_{girl}V_{girl}+M_{ball}V_{ball}=0
 
Nathanael said:
Where did you get V_{girl}=M_{ball}V_{ball}-\frac{M_{ball}}{M_{girl}}?

Use the other equation M_{girl}V_{girl}+M_{ball}V_{ball}=0

(1.8 kg)( 7.5 m/s) / (55kg)
.25 m/s
 
  • #10
Mltn12 said:
(1.8 kg)( 7.5 m/s) / (55kg)
.25 m/s

Yes correct.

Can you do something similar for the other person?
 
  • #11
Nathanael said:
Yes correct.

Can you do something similar for the other person?
MbVb+MballVball=0
(1.8kg)(7.5m/s) / 75
=.18
 
  • #12
Mltn12 said:
MbVb+MballVball=0
(1.8kg)(7.5m/s) / 75
=.18

Yes, that is right.

The reason is that the person slows down the momentum of the ball (by 1.8 kg times 7.5 m/s) and so the persons momentum must increase by the same amount
 
  • #13
So the final velocity for the girl would be .25m/s while the boys would .18 m/s?
 
  • #14
Mltn12 said:
So the final velocity for the girl would be .25m/s while the boys would .18 m/s?

Yep!
 
  • #15
Wow really? It was that simple? I thought there was more to that problem! Thank you so much!
 
  • #16
No problem!

Conservation of momentum is a very useful tool
 
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