Physics Help Forum I can't find my mistake in this zero momentum problem...

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 Nov 5th 2014, 02:48 PM #1 Junior Member   Join Date: Nov 2014 Posts: 3 I can't find my mistake in this zero momentum problem... A body of mass 20m moves at a speed v= 50m/s in the lab frame. It breaks into a pair of fragments, of 4m and 16m, with an energy release of E. This energy appears as extra kinetic energy for the fragments. The fragment 16m moves perpendicular to the original line of flight, measured in the lab frame. (Any change in mass is negligible). Find the velocities of the fragments in the zero momentum frame. I have that the zero momentum frame moves at v. If I say that the speeds of 16m and 4m are p and q respectively, that means that they move at p-v and q-v in the zero momentum frame, doesn't it? I have no idea how to actually find their velocities though...
 Nov 6th 2014, 10:18 AM #2 Junior Member   Join Date: Oct 2014 Posts: 13 Think of the conservation of momentum in the perpendicular direction.
Nov 6th 2014, 12:14 PM   #3
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 Originally Posted by StarsBaby Think of the conservation of momentum in the perpendicular direction.
How do you mean? As in the momentum of the larger particle equals the momentum of the smaller one?

 Nov 6th 2014, 01:46 PM #4 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,324 Assuming the 20m particle is initially moving to the right at 50 m/s, from the point of view of the particle it is stationary and the lab is moving to the left, so its initial momentum is zero. After the explosion the 16m particles is moving upward and to the left (that's how it would appear to go vertical in the lab reference frame), so the 4m particle must be going down and to the right. For the 16m particle, its velocity in the x-direction relative to the zero momentum frame is -50, so its momentum in the x-direction is -50 x 16 = -800. Consequently the 4m particle must have momentum in the x-direction of +800, which means its velocity in the x-direction is 800/4 = 200. We still have unknown y-components of velocity for both, but we know the total KE = E. If we let Vp= vertical velocity of the 16m particle and Vq = vertical velocity of 4m particle, then from vertical momentum: 16Vp + 4Vq = 0 and from energy: (1/2)16(50^2 + Vp^2) + (1/2)4(200^2 + Vq^2) = E So you have two equations in two unknowns. From this you can find the velocities of the two particles relative to the zero momentum frame. Last edited by ChipB; Nov 6th 2014 at 01:51 PM.
Nov 6th 2014, 03:14 PM   #5
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 Originally Posted by ChipB Assuming the 20m particle is initially moving to the right at 50 m/s, from the point of view of the particle it is stationary and the lab is moving to the left, so its initial momentum is zero. After the explosion the 16m particles is moving upward and to the left (that's how it would appear to go vertical in the lab reference frame), so the 4m particle must be going down and to the right. For the 16m particle, its velocity in the x-direction relative to the zero momentum frame is -50, so its momentum in the x-direction is -50 x 16 = -800. Consequently the 4m particle must have momentum in the x-direction of +800, which means its velocity in the x-direction is 800/4 = 200. We still have unknown y-components of velocity for both, but we know the total KE = E. If we let Vp= vertical velocity of the 16m particle and Vq = vertical velocity of 4m particle, then from vertical momentum: 16Vp + 4Vq = 0 and from energy: (1/2)16(50^2 + Vp^2) + (1/2)4(200^2 + Vq^2) = E So you have two equations in two unknowns. From this you can find the velocities of the two particles relative to the zero momentum frame.

Thank you for this! I'm just wondering; why does KE=E?

Nov 7th 2014, 07:48 AM   #6
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 Originally Posted by Bmurw I'm just wondering; why does KE=E?
Where else can the energy E go to other than becoming kinetic energy of the system?

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