Question

shell of mass 5m acted upon by no external force initially at rest bus into three fragments of masses M 2m and 2m respectively the first to fragment move in opposite directions with velocities of magnitude 2v and 3V respectively the third fragment velocity will be in which direction ​

Answer 1

Answer:

2v, in the direction, the first fragment of mass m

Explanation:

Mass of the shell = 5m

The shell implodes into 4 fragments of mass m, 2m and 2m

∵ the first two fragments move in the opposite direction with velocity 2v and 3v

∴ Assuming that the fragments move in the +x and -x directions respectively, the velocity vectors can be written as

velocity of fragment of mass m = $2v\hat{i}$

velocity of fragment of mass 2m = $-3v\hat{i}$

Let the velocity vector of the third fragment of mass 2m is $\vec{v}$

The initial velocity of the shell is 0

Therefore, applying the conservation of linear momentum principle

$5m\times0=m2v\hat{i}+2m(-3v\hat{i})+2m\vec{v}$

or, $0=2mv\hat{i}-6mv\hat{i}+2m\vec{v}$

or, $2m\vec{v}=4mv\hat{i}$

or, $\vec{v}=2v\hat{i}$

Thus, the velocity of the third fragment will be 2v in the direction of the first fragment

Answer 2

Answer:v = 0

Explanation:

Let the velocity of third fragment is v

Then by the conservation of linear momentum:

5M×0=M×2v−2M×v+2M×v

or, v =0

So, the third fragment will be at rest.