Help me find the final velocity tnx

[chel]

0.5Kg snowball moving at 20m/s strikes and sticks to a 70Kg man standing on the frictionless surface of a frozen pond. what is the final velocity?

 

[chel]

help.. please.. :(

 

[Doc Al]

Show what you've done so far and where you are stuck.

Hint: What's conserved?

 

[chel]

uhmm, there was no conservation happened because there was an external force coming from the snowball i think... :(

 

[Doc Al]

Try considering the snowball and man as a single system.

 

[chel]

uhmm,, will this equation solve my problem?

(0.5kg) * (20m/s)= (70kg)* (?)

im not really sure if I am going on the right track i don't know much about physics..
i feel so helpless by not knowing these things.. :(

 

[Cyosis]

It's close, but not correct. Use conservation of momentum. The system here is the man and the snowball. What is the momentum before the collision? What is the momentum after the collision? To elaborate a bit further. Before the impact the snow ball has a certain speed and mass and so does the man, his speed is 0 since he's at rest. The question states that after the collision the snowball sticks to the man. Can you do anything with this information?

 

[chel]

uhmm, the momentum before and after the collision was zero?

 

[Doc Al]

Another hint: This is an example of a totally inelastic collision. This short reference might help: https://www.physicsforums.com/showpost.php?p=2157983&postcount=5"

 

[chel]

uhmm, the momentum before and after the collision is zero??

 

[Doc Al]

uhmm, the momentum before and after the collision was zero?

No. To find the total momentum of the system, just add up the momentum of each part. What's the initial momentum of the snowball? Of the man?

 

[Cyosis]

uhmm, the momentum before and after the collision was zero?

In the system of man and snowball total momentum is conserved. The total momentum before the collision is not zero, thus it cannot be zero after the collision. Could you write down the equations for momentum for the system before the collision? p=...?

Edit: Since I didn't post any new information with this post I will expand a bit upon it.

Before the collision:
What is the mass of the snowball?
What is the speed of the snowball?
What is he mass of the man?
What is the speed of the man?

After the collision:
What is the mass of the man and the snowball that is stuck to him?
What is the speed of the combined object that is man and snowball?

Now what is the definition of momentum? Plug in the numbers and solve.

Answer all these questions if you haven't found the solution yet.

 

[chel]

can i use this formula?
(m)*(20m/s)+(m)(0m/s)=(2m)(velocity after the collision)?

 

[Cyosis]

All the masses are the same in that equation, while in your problem they are not. Instead of just using a formula you don't understand, try to understand what's going on. When you do that you will see you will be able to solve all other problems like this easily without having to worry about what formula to use.

Can you tell us what conservation of momentum means and do you know how you write momentum down in equation form?

 

[chel]

can i use this formula?

(0.5kg)*(20m/s)+(70kg)(0m/s)=(0.5kg+70kg)(Vafter)?
10kg.m/s=(70.5kg)(Vafter)
Vafter=10kg.m/s / 70.5kg
= 0.14m/s

 

[Cyosis]

You got it!

 

[chel]

accdg to what i have read, the momentum is unchanged if there is no external force acted on the system.. the energy is conserved...

 

[chel]

hai,, guys thanks for the effort in helping me solving this problem..

 

[chel]

you mean to say tama ako? i mean i got it the correct answer?

 

[Cyosis]

Tama ako? Anyway yes you found the correct method and answer to your problem.

 

[chel]

(0.5kg) * (20m/s)= (70kg)* (?) ==> this formula gave me same answer.. what made this wrong i used analysis using this formula.. you said this is close but wrong? hehehe.. but I am happy i got it correct

 

[Cyosis]

It's a matter of significant digits really.

The correct equation to use was [itex]m_{sb} v_{sb}+m_{man}v_{man}=(m_{sb}+m_{man}) v_{final}[/itex]. The interesting part is [itex](m_{sb}+m_{man}) =70.5kg[/itex].

The equation you used the first time was [itex]m_{sb} v_{sb}+m_{man}v_{man}=m_{man} v_{final}[/itex]. Here you are ignoring that when a snowball is stuck to a man his total mass is increased. Because the snowball is so light both equations gave the same answer with a 2-decimal accuracy. Try to recalculate it by saying that the mass of the snowball is 100kg, keep the rest of the numbers the same. You will see that your answer is very different.

More intuitively: If I throw a snowball at you, you won't move much because its mass is low. If I were to throw a wrecking ball at you, you would move quite a bit since its mass is high compared to yours, assuming both snowball and wrecking ball hit you at the same speed.

 

[chel]

Tama ako? Anyway yes you found the correct method and answer to your problem.

tama ako means I am correct... bye.. thanks :zzz:

 

[chel]

It's a matter of significant digits really.

The correct equation to use was [itex]m_{sb} v_{sb}+m_{man}v_{man}=(m_{sb}+m_{man}) v_{final}[/itex]. The interesting part is [itex](m_{sb}+m_{man}) =70.5kg[/itex].

The equation you used the first time was [itex]m_{sb} v_{sb}+m_{man}v_{man}=m_{man} v_{final}[/itex]. Here you are ignoring that when a snowball is stuck to a man his total mass is increased. Because the snowball is so light both equations gave the same answer with a 2-decimal accuracy. Try to recalculate it by saying that the mass of the snowball is 100kg, keep the rest of the numbers the same. You will see that your answer is very different.

More intuitively: If I throw a snowball at you, you won't move much because its mass is low. If I were to throw a wrecking ball at you, you would move quite a bit since its mass is high compared to yours, assuming both snowball and wrecking ball hit you at the same speed.

oh! i see,, i'll keep that in mind... well thanks anyway,, i still have 10 more questions to answer tomorrow,, thank you,, nyters and byers