Conservation law of momentum.... Tricky question

[Mrchilko]

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

P_a + P_b = 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]

 

[haruspex]

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

P_a + P_b = 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.

 

[Mrchilko]

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..

 

[FreezingFire]

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

P_a + P_b = 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".

 

[haruspex]

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.

 

[Mrchilko]

Yes 2.5m/s is only the velocity and the momentum has the units of kg*m/s

 

[haruspex]

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?

 

[Mrchilko]

Mass... So it would look like (3.5m/s)(?) + (0m/s)(5kg) = (5+?kg) + ( 2.5m/s)

 

[haruspex]

Mass... So it would look like (3.5m/s)(?) + (0m/s)(5kg) = (5+?kg) + ( 2.5m/s)

Kg+m/s?!

 

[Mrchilko]

M*v

 

[haruspex]

M*v

Ok, so write out the equation correctly and solve it.