How can I find the mass of a junior curling stone using momentum?

[The_big_dill]

Homework Statement

I am doing a laboratory on momentum with curling stones, and i know the mass of a regular curling stone, but not the mass of the junior stone (which is what i need to find out).

A regular curling stone was pushed into a junior one with the following data:

NOTE: All accelerations are assumed to affect all stones the same, mass of regular curling stone is 20Kg

Initial

Reg Stone:
$$\Delta$$d = 6.5m
$$\Delta$$t = 3.3s
a = -0.13m/s²

Final

Reg Stone:
$$\Delta$$d = 1.82m (cos40)
a = -0.13m/s²
V₂ = 0

junior Stone:
$$\Delta$$d = 11.1m(cos30)
a = -0.13m/s²
V₂ = 0

Homework Equations

$$\Delta$$d = V₁$$\Delta$$t+1/2a$$\Delta$$t²

V₂² = V₁² + 2a$$\Delta$$d

P_T = P_T'

The Attempt at a Solution

Using ($$\Delta$$d = V₁$$\Delta$$t+1/2a$$\Delta$$t²) i found the initial velocity of the regular stone to be 2.18m/s

Using (V₂² = V₁² + 2a$$\Delta$$d) i found the final velocity of both stones after impact to be

V₁' = 0.6m/s

V₂' (junior stone) = 1.58m/s

P_T = P_T'

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Since V₂ = 0 (junior stone initially not moving), m₂v₂ Can be eliminated.

m₁v₁ = m₁v₁' + m₂v₂'

rearrange and solve for m₂

m₂ = m₁ (v₁ - v₁')/v₂'

m₂ = 20Kg <<<< not possible... ^^^^what am i doing wrong?

 

[soothsayer]

How did you find the final velocity of both stones after impact?

 

[soothsayer]

Oh, nevermind, I see your problem

To solve it, remember that the velocity at which the regular curling stone collides with the junior stone (the momentum that is conserved) is NOT equal to the initial velocity of the regular curling stone, 2.18. You must factor in the acceleration of the regular stone between the time it starts at 2.18m/s and the moment it collides with the junior stone.

v₂=v₁ + at
where v₁ is 2.18m/s, not v₂

Then use the velocity of the regular stone right before collision (v₂ in the above equation) as the v₁ for the momentum conservation equation you have, and you'll come up with a reasonable value for mass of the junior stone.

 

[gneill]

Reg Stone:
$$\Delta$$d = 1.82m (cos40)
a = -0.13m/s²
V₂ = 0

junior Stone:
$$\Delta$$d = 11.1m(cos30)
a = -0.13m/s²
V₂ = 0

I'm unfamiliar with the notation used for the distances. What does the "cos40" and "cos30" imply? Does it mean that only the x-axis component of the distance is being specified (the implication being that the the total displacement is $$\Delta$$d/cos40)?

 

[The_big_dill]

Oh, nevermind, I see your problem
v₂=v₁ + at
where v₁ is 2.18m/s, not v₂

I appreciate your help.

Are you sure you are not missing a variable (or using the correct ones) in that equation? because i have never seen that one before, although ones that are similar from the equations of uniform acceleration.

 

[The_big_dill]

I'm unfamiliar with the notation used for the distances. What does the "cos40" and "cos30" imply? Does it mean that only the x-axis component of the distance is being specified (the implication being that the the total displacement is $$\Delta$$d/cos40)?

It is the angle from the horizontal (that being a straight line along the ice)

 

[soothsayer]

It's a basic equation, it just shows that change in velocity (v2-v1) = acceleration x time (amount of time object is accelerating)

For example, if a car accelerates from rest at 4m/s² for 3 seconds, what is the change in velocity? 4 x 3 = 12m/s, so the car will be going 12m/s after 3 seconds of acceleration of 4m/s²

 

[gneill]

It is the angle from the horizontal (that being a straight line along the ice)

Sorry if I'm being thick here, but I understand that there is an angle implied. What I want to know is if it implies anything else about the distance number associated with it.

Suppose I specify a distance as R cos50. Is R the magnitude of the distance vector, or only the x-axis component (making the magnitude of the vector R/cos(40))?

 

[soothsayer]

gneill, judging by the numbers he got for the velocities of the stones after the collision, he seems to have multiplied... d=Rcos(40), not d=R or d=R/cos(40). I too was confused by the notation.

 

[The_big_dill]

gneill, judging by the numbers he got for the velocities of the stones after the collision, he seems to have multiplied... d=Rcos(40), not d=R or d=R/cos(40). I too was confused by the notation.

apologize for the confusion, soothsayer is correct.

 

[The_big_dill]

It's a basic equation, it just shows that change in velocity (v2-v1) = acceleration x time (amount of time object is accelerating)

For example, if a car accelerates from rest at 4m/s² for 3 seconds, what is the change in velocity? 4 x 3 = 12m/s, so the car will be going 12m/s after 3 seconds of acceleration of 4m/s²

Sorry, i guess i had a brain fart, i am use to seeing it as a=v/t

 

[gneill]

gneill, judging by the numbers he got for the velocities of the stones after the collision, he seems to have multiplied... d=Rcos(40), not d=R or d=R/cos(40). I too was confused by the notation.

Hmm. Two problems. The frictional force that's providing the deceleration that stops the stones should be working along the line of motion, not just the x-component, and both post impact velocities apparently are associated with positive angles (both stones move along trajectories that are above the horizontal after the collision). Conservation of momentum would then indicate that the original stone was not moving "horizontally", yet no angle was supplied.

 

[The_big_dill]

Hmm. Two problems. The frictional force that's providing the deceleration that stops the stones should be working along the line of motion, not just the x-component, and both post impact velocities apparently are associated with positive angles (both stones move along trajectories that are above the horizontal after the collision). Conservation of momentum would then indicate that the original stone was not moving "horizontally", yet no angle was supplied.

Let me clarify, in my original post i was attempting to make everything more brief so it is less discouraging to read.

The regular stone hit the junior one at an angle, making one go 40 degrees and the other go -30 degrees, if that makes sense.

I calculated a deceleration that would apply to all stones from that point on to be a=-0.13

 

[gneill]

Let me clarify, in my original post i was attempting to make everything more brief so it is less discouraging to read.

The regular stone hit the junior one at an angle, making one go 40 degrees and the other go -30 degrees, if that makes sense.

I calculated a deceleration that would apply to all stones from that point on to be a=-0.13

These angles are with respect to what axis? The initial direction of travel of the Regular Stone? If so, I think that there's a problem with the data. Momentum must be conserved in both the X and Y directions. Using the angles given and post-impact velocities deduced, the Y-direction momentum is not conserved, as it ends up being non-zero.

 

[The_big_dill]

These angles are with respect to what axis? The initial direction of travel of the Regular Stone? If so, I think that there's a problem with the data. Momentum must be conserved in both the X and Y directions. Using the angles given and post-impact velocities deduced, the Y-direction momentum is not conserved, as it ends up being non-zero.

i don't know if i am missing something... but i could simplify it by just finding what $$\Delta$$d*Cos$$\theta$$ is. i am simply finding the X-component, i don't think i need both components, all i need is the distance traveled in a straight line.

 

[gneill]

i don't know if i am missing something... but i could simplify it by just finding what $$\Delta$$d*Cos$$\theta$$ is. i am simply finding the X-component, i don't think i need both components, all i need is the distance traveled in a straight line.

You wrote for your conservation of momentum equation,

m1v1 = m1v1' + m2v2'

and then proceeded to deal with the x-components on the RHS. If the Regular Stone was not actually traveling along the x-axis when it hit the Junior Stone, it too will have x and y components to deal with; the velocity that it has at the moment of impact will have x and y components. So there is some unknown angle associated with the Regular Stone's initial velocity, and the x-component of this velocity will be some fraction of the speed of the stone.

That this is so is clear from the fact that the Y-momenta of the two stones immediately after impact do not sum to zero when you assume that Regular Stone traveled directly along the x-axis.

 

[soothsayer]

You wrote for your conservation of momentum equation,

m1v1 = m1v1' + m2v2'

and then proceeded to deal with the x-components on the RHS. If the Regular Stone was not actually traveling along the x-axis when it hit the Junior Stone, it too will have x and y components to deal with; the velocity that it has at the moment of impact will have x and y components. So there is some unknown angle associated with the Regular Stone's initial velocity, and the x-component of this velocity will be some fraction of the speed of the stone.

That this is so is clear from the fact that the Y-momenta of the two stones immediately after impact do not sum to zero when you assume that Regular Stone traveled directly along the x-axis.

How do you know that the y-momenta of the stones did not sum to zero if you don't know the mass of the junior stone?

Also, I felt it was implied that the angle given of the stones after the collision was with respect to the direction the regular stone was traveling before the collision.

 

[soothsayer]

m₁v₁ = m₁v₁' + m₂v₂'

v₁ = 2.18 + at = 2.18 + (-0.13)(3.3) = 1.75_m/s (velocity the moment before the collision)

using the formula: v₂² = v₁² + 2ad, we can focus on only the component of velocity in the original direction of the curling stone's travel by determining d₂ = 11.1cos(30) = 9.61
and d₁ = 1.82cos(40) = 1.39
(I guess x would be more appropriate than d here)
we come up with the values that the big dill came up with for the velocities after the collision in the OP:
v₁' = 0.6_m/s
v₂' = 1.58_m/s

We plug these into the momentum conservation equation I posted above, this is where the big dill went wrong, he plugged in 2.18_m/s for v₁ which was the initial velocity of the regular stone INSTEAD of 1.75_m/s for velocity of the regular stone instantaneously before colliding with the junior stone, as he should have (as I demonstrated above, accounting for deceleration).

Now, if you plug in all the correct values, the momentum conservation equation becomes:

(20)(1.75) = (20)(0.6) + 1.58m₂
Solving for m₂ we get 1.85_kg

That should be the mass of the junior stone. To check, let's plug these values into kinetic energy conservation and see if they match up:

(1/2)m₁v₁² = (1/2)m₁v₁'² + (1/2)m₂v₂'²

here, energy is a scalar, so we don't want to deal with velocity components, only the full speed of the stones. I'll spare you the derivations I did and say
v₂'² = 2.88
v₁'² = 0.473

Plug these values in, to the above equation and you'll see the two sides do NOT equal each other, this means either I made a mistake (not too unlikely) or there's something wrong with the problem, I'd look at those weird trig components that still confuse me. I'm going to try it again without factoring those in and tell you the results.

EDIT: Ran the numbers, it still does not match up. Perhaps the collision is supposed to be inelastic?

 

[gneill]

To account for the initial velocity of the Regular stone being at some undetermined angle with respect to the assumed x-axis, or equivalently, the Regular stone striking the Junior stone off-center, divide the Regular stone's velocity into components as usual.

The methods used to find the initial speeds of the stones pre- and post-impact are fine.

To summarize the plot so far:

Regular Stone:
   m1 = 20 kg
   vr1 = 1.755 m/s        [tex]\phi[/tex]  = ? (Unknown)
   vr2 = 0.688 m/s        [tex]\theta[/tex]r = 40 degrees

Junior Stone:
   m2 = ?
   vj1 = 0  (at rest)
   vj2 = 1.699 m/s        [tex]\theta[/tex]j = 30 degrees

Write the 2D conservation of momentum equations:

   m1*vr1*cos([tex]\phi[/tex]) = m1*vr2*cos([tex]\theta[/tex]r) + m2*vj2*cos([tex]\theta[/tex]j)
   m1*vr1*sin([tex]\phi[/tex]) = m1*vr2*sin([tex]\theta[/tex]r) + m2*vj2*sin([tex]\theta[/tex]j)

Solve these simultaneous equations for m2.

When I solve the equations I come up with

m2 = 16.44 kg
$$\phi$$ = -8.39 degrees

Momentum is conserved in all directions, and the mass of the Junior stone even looks reasonable.