Exploding Mass Problem: Finding Distance Between Two Pieces After Landing

[issacnewton]

here is a problem i am doing

consider a mass which is freely falling vertically down. neglect air resistance. at some point this mass explodes and is divided in two pieces. mass m₁ and mass m₂. these pieces fall on the ground at the same time. the first piece falls at a distance s from the place it would have landed (straight down) had the explosion not occurred. please find the distance between the pieces after they land on the ground.

here is my attempt. the exploding pieces can have x and y components of the velocities. but we need not consider the y components of the velocities since the linear momentum in y direction is not conserved. if v₁ and v₂ are the velocities in the
x direction , we can write using conservation of linear momentum.

$$m_1 v_1=m_2 v_2$$

Now $s=v_1 t$.let s' be the horizontal distance covered by the second piece, then
$s' = v_2 t$. The distance between the pieces then is s+s'.

$$s+s'=v_1 t + v_2 t$$

$$s+s'= \left(1+\frac{m_1}{m_2}\right)v_1 t$$

$$s+s'= \left(1+\frac{m_1}{m_2}\right)s$$

does it sound right ?

 

[gneill]

here is a problem i am doing

consider a mass which is freely falling vertically down. neglect air resistance. at some point this mass explodes and is divided in two pieces. mass m₁ and mass m₂. these pieces fall on the ground at the same time. the first piece falls at a distance s from the place it would have landed (straight down) had the explosion not occurred. please find the distance between the pieces after they land on the ground.

here is my attempt. the exploding pieces can have x and y components of the velocities. but we need not consider the y components of the velocities since the linear momentum in y direction is not conserved.

Since you haven't specified a coordinate system it's hard to judge the truth of this statement for the present case. Momentum is always conserved unless an external force is acting. Are you taking the Y-direction to be the vertical direction?

if v₁ and v₂ are the velocities in the
x direction , we can write using conservation of linear momentum.

$$m_1 v_1=m_2 v_2$$

Now $s=v_1 t$.let s' be the horizontal distance covered by the second piece, then
$s' = v_2 t$. The distance between the pieces then is s+s'.

$$s+s'=v_1 t + v_2 t$$

$$s+s'= \left(1+\frac{m_1}{m_2}\right)v_1 t$$

$$s+s'= \left(1+\frac{m_1}{m_2}\right)s$$

does it sound right ?

Looks good.

You could also have used the fact that the trajectory of the center of mass is unaffected by the explosion, so the center of mass should remain at (0,0) on the ground.

 

[issacnewton]

thanks gneill. yes y is taken as vertical direction... yes CM just comes down straight. but how can we use that fact to solve the problem

 

[gneill]

thanks gneill. yes y is taken as vertical direction... yes CM just comes down straight. but how can we use that fact to solve the problem

If the center of mass is known (well, defined to be at the origin) and there are two pieces of known mass, one of which has a known distance from the center of mass, it should be simple enough to place the other so that the center of mass conditions hold.

 

[issacnewton]

ok makes sense

 

[ehild]

The explosion can be considered instantaneous, that is the vertical components of the momentum are not affected by gravity during such a short time: Both the vertical and horizontal components will conserved.
But you have the information that both pieces fell to the ground at the same time: Therefore the pieces did not change their y component of velocity.
If the vertical components were different after explosion, they would not arrive at the ground at the same time. When one piece has fallen, conservation of momentum would not hold any more, as the ground means an extra force. In that case, you should find the landing point of both pieces.

ehild

 

[issacnewton]

thanks ehild...