Conservation law of momentum... Tricky question

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Homework Statement


A ball rolls at a velocity of 3.5m/s toward a 5.0kg ball at rest. They collide and stick and move off at 2.5 m/s. What was the mass of the moving ball?[/B]


Homework Equations


Pa + Pb = 0
M(a)V(a) = - M(b)V(b)[/B]


The Attempt at a Solution


So.. The ball at rest had zero momentum... Therefore how are we supposed to find the momentum in total if we are only given 3 numbers to work w/ ... We can't determine the moving ball momentum.. Soo[/B]
 

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  • #2
haruspex
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Homework Statement


A ball rolls at a velocity of 3.5m/s toward a 5.0kg ball at rest. They collide and stick and move off at 2.5 m/s. What was the mass of the moving ball?[/B]


Homework Equations


Pa + Pb = 0
M(a)V(a) = - M(b)V(b)[/B]


The Attempt at a Solution


So.. The ball at rest had zero momentum... Therefore how are we supposed to find the momentum in total if we are only given 3 numbers to work w/ ... We can't determine the moving ball momentum.. Soo[/B]
You are given four velocities and one mass. There's only one unknown, and you have an equation.
 
  • #3
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But .... If the second ball at rest has a momentum of 0kg°m/s .... Then M(a)3.5m/s (a)= 5.0kg(b)0m/s(b)

So saying (?)*(3.5m/s) + (5.0kg)*( 0 m/s) = 2.5 m/s..
 
  • #4

Homework Statement


A ball rolls at a velocity of 3.5m/s toward a 5.0kg ball at rest. They collide and stick and move off at 2.5 m/s. What was the mass of the moving ball?[/B]


Homework Equations


Pa + Pb = 0
M(a)V(a) = - M(b)V(b)[/B]


The Attempt at a Solution


So.. The ball at rest had zero momentum... Therefore how are we supposed to find the momentum in total if we are only given 3 numbers to work w/ ... We can't determine the moving ball momentum.. Soo[/B]
Check the relevant equation ##P_a + P_b = 0##. Are you sure that's correct? Shouldn't it be something else:
$$P_i = P_f \implies P_{a_i} + P_{b_i} = P_{a_f} + P_{b_f}$$
The ##i## denotes "initial" and ##f## denotes "final".
 
  • #5
haruspex
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But .... If the second ball at rest has a momentum of 0kg°m/s .... Then M(a)3.5m/s (a)= 5.0kg(b)0m/s(b)

So saying (?)*(3.5m/s) + (5.0kg)*( 0 m/s) = 2.5 m/s..
The term on the right has the wrong dimension. You've left out a factor.
 
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Yes 2.5m/s is only the velocity and the momentum has the units of kg*m/s
 
  • #7
haruspex
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Yes 2.5m/s is only the velocity and the momentum has the units of kg*m/s
So correct it. What factor did you leave out?
 
  • #8
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Mass... So it would look like (3.5m/s)(?) + (0m/s)(5kg) = (5+?kg) + ( 2.5m/s)
 
  • #9
haruspex
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Mass... So it would look like (3.5m/s)(?) + (0m/s)(5kg) = (5+?kg) + ( 2.5m/s)
Kg+m/s?!
 
  • #11
haruspex
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M*v
Ok, so write out the equation correctly and solve it.
 

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