Tracking the Third Piece of an Exploding Ball

[Student91]

A ball of mass m at rest at the coordinate origin exploded into three equal pieces.
At a certain instant, one piece is on the x-axis at x=40cm, and another is at x=20cm, y=-60cm. where is the third piece at that instant


I don't have an idea how to solve it I tried many time but I don't reach an answer.

I have to use F_ex=Ma_c.m

 

[Stonebridge]

It is solved using conservation of momentum; or centre of mass considerations. (It amounts to the same thing)
As there are no external forces acting, the centre of mass of the combined 3 fragments cannot have changed position, and must still be at the centre of the coordinate system.
Alternatively, if you let the time interval = t, you can write down the x and y coordinates of the momentum of the 2 known fragments. (You have the distance traveled in that time from the coordinates. Velocity = distance/time)
The total momentum of those two plus the unknown fragment must equal zero.

I don't see how you can use F=ma here.

 

[Student91]

My Physics teacher told me it solved by F=ma ..

 

[Borek]

Sorry to hear that.

Momentum conservation it is.

 

[Student91]

Is the answer going to be -60i+60j ?

 

[Stonebridge]

What do i and j represent here? Are they unit cm vectors in the x and y directions?
You would normally give the answer as x=-60cm and y=+60cm
How have you calculated those values. You would need to show some working, not a guess.

 

[Student91]

My Work :​

By Conservation of Momentum :

P_i = P_f
0= mv₁ + mv₂ + mv₃

Devid By 3m ..

= v₁ + v₂ + v₃
= d₁/3t + d₂/3t + d₃/3t

Devide By 3t ..

d₃ = -( d₁ + d₂ )

d₃ = -( 40i + 20i - 60j )

d₃ = -40i - 20i + 60j

d₃ = -60i + 60j

So x= -60
and y = 60

 

[Student91]

Up !

 

[Stonebridge]

Yes those answers are correct. (In centimeters)