Conservation of Momentum in an Explosion

[Priyadarshini]

Homework Statement

During an explosion, a bomb explodes into 3 pieces. Two fragments, whose masses are 0.8kg and 0.5kg fly off with velocities of 10m/s and 16m/s respectively along the paths at right angles to each other. If the third fragment goes off with a velocity of 24m/s, then find its mass and direction w.r.t the first fragment.

Homework Equations

- Components ( u sin A and u cos A)
- Conservation of momentum. The horizontal components=0 and the vertical components=0

The Attempt at a Solution

For M1
horizontal component of velocity= 10 cos A
vertical component of velocity= 10 sin A
Horizontal momentum= 0.8 x 10cosA
= 8cosA
Vertical momentum=10sinA x 0.8
= 8sinA

For M2
Horizontal component of velocity=16sinA
Vertical Component of velocity= 16cosA
Horizontal momentum= 8sinA
Vertical momentum= 8cosA

How do I find out the direction in which the third pieces moves in? Without the direction on the diagram I've drawn, I cannot split its velocity in component.
Thanks in advance!

 

[Bystander]

You'll need to either include your diagram, or specify either "clock" or compass directions for us.

 

[jbriggs444]

How do I find out the direction in which the third pieces moves in?

Momentum is conserved, right? And you know the final momentum of the first two pieces. If you assume that the bomb started at rest, what must be the final momentum of the third piece?

 

[Ellispson]

Homework Statement

During an explosion, a bomb explodes into 3 pieces. Two fragments, whose masses are 0.8kg and 0.5kg fly off with velocities of 10m/s and 16m/s respectively along the paths at right angles to each other. If the third fragment goes off with a velocity of 24m/s, then find its mass and direction w.r.t the first fragment.

Homework Equations

- Components ( u sin A and u cos A)
- Conservation of momentum. The horizontal components=0 and the vertical components=0

The Attempt at a Solution

For M1
horizontal component of velocity= 10 cos A
vertical component of velocity= 10 sin A
Horizontal momentum= 0.8 x 10cosA
= 8cosA
Vertical momentum=10sinA x 0.8
= 8sinA

For M2
Horizontal component of velocity=16sinA
Vertical Component of velocity= 16cosA
Horizontal momentum= 8sinA
Vertical momentum= 8cosA

How do I find out the direction in which the third pieces moves in? Without the direction on the diagram I've drawn, I cannot split its velocity in component.
Thanks in advance!

Since momentum is being conserved.The momentum of the third particle will be anti parallel to the resultant of the momenta of the first and second particles.

 

[Priyadarshini]

Since momentum is being conserved.The momentum of the third particle will be anti parallel to the resultant of the momenta of the first and second particles.

But why is the momentum anti parallel?

 

[Priyadarshini]

Here's the diagram.

 

[Dr. Courtney]

3D or 2D?

 

[Ellispson]

But why is the momentum anti parallel?

Because the bomb is initially at rest with a total momentum of zero.The momentum after the explosion should be zero too because there are no external forces acting on it.

 

[Priyadarshini]

3D or 2D?

The question doesn't say. But I think it's 2D.

 

[Priyadarshini]

Because the bomb is initially at rest with a total momentum of zero.The momentum after the explosion should be zero too because there are no external forces acting on it.

Thank you! I get it now!