Question

22. (a) A bomb of mass 5kg explodes into two equal fragments. At what angl
these fragments fly apart?
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Answer 1

Answer:

Let the bomb have a mass 2m,

Let the mass and velocity of the first fragment be m and v1 and of the 2nd be m and v2 respectively.

The initial velocity of the bomb be 0 assuming the bomb is at rest in air when it exploding

Intial momentum = final momentum

2m(0) = mv1 + mv2

=> mv1 + mv2 = 0

=> v1 = -v2

This means the two fragments will have equal  but opposite velocities.

Thus direction of fragments opposite of each other.

Answer 2

Given : Mass of bomb = 5 kg

            Bomb exploded into two equal fragments

To Find : Angle at which two equal fragment fly apart.

Solution :

Initially the momentum of bomb is zero. So according to the law of conservation of linear momentum the final momentum of the system must remain zero.

As initial bomb mass = 5kg and bomb exploded in two equal parts ,so mass of first frangment and 2nd fragment is

$\[{m_1}\]$ = 2.5 kg

$\[{m_2}\]$= 2.5 kg

 Let the velocity of the first fragment v1 and of the 2nd fragment be v2.

The initial velocity of the bomb be 0 assuming the bomb is at rest in air when it exploding.

According to the Law of conservation of linear momentum.

Initial momentum = Final momentum.

$\[{m_1}\]$$\[{u_1}\]$ + $\[{m_2\\}\]$$\[{u_2}\]$u = $\[{m_1}{v_1} + {m_2}{v_2}\]$

2*5*(0) = 2.5*V1 + 2.5*v2

=> 2.5($\[{v_1}\]$+$\[{v_2}\]$) = 0

=> $\[{v_1}\]$+$\[{v_2}\]$ = 0

=>  $\[{v_1}\]$ = -$\[{v_2}\]$

Thus the two fragments will have equal but opposite velocities.

Hence, Angle at which two equal fragment fly apart is $\[180^\circ \]$, opposite to each other.