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

In summary: Keep practicing and you'll get the hang of it. Good job!In summary, the problem involves a 55-kg girl standing near and to the left of a 75-kg boy on a frictionless frozen pond. The girl throws a 1.8-kg ice ball horizontally at 7.5 m/s and the boy catches it. Using the principles of conservation of momentum, we can determine that the final velocities of the boy and girl would be 0.18 m/s and 0.25 m/s respectively.
  • #1
Mltn12
8
0

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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  • #2
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)
 
  • #3
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 !
 
  • #4
V=Vo+at?
A=t/v, but I don't have t.
 
  • #5
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.
 
  • #6
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?
 
  • #7
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
 
  • #8
Mltn12 said:
Vfg=MballVb-Mb / Mg

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

Where did you get [itex]V_{girl}=M_{ball}V_{ball}-\frac{M_{ball}}{M_{girl}}[/itex]?

Use the other equation [itex]M_{girl}V_{girl}+M_{ball}V_{ball}=0[/itex]
 
  • #9
Nathanael said:
Where did you get [itex]V_{girl}=M_{ball}V_{ball}-\frac{M_{ball}}{M_{girl}}[/itex]?

Use the other equation [itex]M_{girl}V_{girl}+M_{ball}V_{ball}=0[/itex]

(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
 

1. How do you determine the forces acting on the boy and girl on the frozen pond?

To determine the forces acting on the boy and girl, we can use Newton's Laws of Motion. First, we must identify all the external forces acting on the objects, such as the normal force from the ice and the force of gravity. Then, we can use vector addition to determine the net force on the objects.

2. How does the friction between the ice and the skates affect the motion of the boy and girl?

The friction between the ice and the skates will provide a force that opposes the motion of the boy and girl. This force, known as the kinetic friction force, will decrease the speed of the objects as they move across the ice. The amount of friction will depend on the coefficient of friction between the ice and the skates.

3. Can you use conservation of energy to solve this physics problem?

Yes, conservation of energy can be used to solve this problem. The total energy of the system, which includes the kinetic energy of the boy and girl and the potential energy of the ice, will remain constant throughout the motion. By setting the initial and final energies equal to each other, we can solve for the velocity of the objects.

4. How do you take into account the mass of the boy and girl when solving this problem?

The mass of the boy and girl will affect the forces acting on them and their acceleration. According to Newton's Second Law of Motion, the net force on an object is equal to its mass multiplied by its acceleration. Therefore, we must consider the mass of the objects when determining the net force and solving the equations of motion.

5. Can air resistance be ignored in this problem?

In most cases, air resistance can be ignored in this problem because the objects are moving on a horizontal surface and the distance they travel is relatively short. However, if the problem involves objects moving at high speeds or over longer distances, the effects of air resistance may need to be taken into account.

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