Question

A giant "egg" explodes as part of a fireworks display. The egg is at rest before...

A giant "egg" explodes as part of a fireworks display. The egg is at rest before the explosion, and after the explosion, it breaks into two pieces, with piece B moving in the positive x direction. The masses of both pieces are indicated in (Figure 1) , shown traveling in opposite directions.

The component of the momentum of piece B, pBx,f, is measured to be +500 kg?m/s after the explosion. Find the component of the momentum pAx,f of piece A after the explosion.

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Answer #1
Concepts and reason

The main concept used to solve the problem is law of conservation of momentum.

Use law of conservation of momentum to calculate the momentum of the piece A after the explosion.

Fundamentals

The conservation of momentum states that the initial momentum is equal to the final momentum. It is expressed as follows:

pe=pAf+pBf{\vec p_{\rm{e}}} = {\vec p_{{\rm{Af}}}} + {\vec p_{{\rm{Bf}}}}

Here, pe{\vec p_{\rm{e}}} is the momentum of the giant egg before explosion, pAf{\vec p_{{\rm{Af}}}} is the final momentum of piece A and pBf{\vec p_{{\rm{Bf}}}} is the final momentum of piece B.

According to the law of conservation of momentum,

pe=pAf+pBf{\vec p_{\rm{e}}} = {\vec p_{{\rm{Af}}}} + {\vec p_{{\rm{Bf}}}}

The egg is at rest before the explosion, so the momentum must be zero.

Substitute 0kgm/s0{\rm{ kg}} \cdot {\rm{m/s}} for pe{\vec p_{\rm{e}}} in above equation as follows:

0kgm/s=pAf+pBfpAf=pBf\begin{array}{c}\\0{\rm{ kg}} \cdot {\rm{m/s}} = {{\vec p}_{{\rm{Af}}}} + {{\vec p}_{{\rm{Bf}}}}\\\\{{\vec p}_{{\rm{Af}}}} = - {{\vec p}_{{\rm{Bf}}}}\\\end{array}

The piece B is moving along positive x- direction. Hence, the momentum of the piece B is given by,

pBf=500kgm/s(x^){\vec p_{{\rm{Bf}}}} = 500{\rm{ kg}} \cdot {\rm{m/s}}\left( {\hat x} \right)

Substitute 500kgm/s(x^)500{\rm{ kg}} \cdot {\rm{m/s}}\left( {\hat x} \right) for pBf{\vec p_{{\rm{Bf}}}} in equation pAf=pBf{\vec p_{{\rm{Af}}}} = - {\vec p_{{\rm{Bf}}}} for pBf{\vec p_{{\rm{Bf}}}} in above equation as follows:

pAf=500kgm/s(x^)=500kgm/s(x^)\begin{array}{c}\\{{\vec p}_{{\rm{Af}}}} = - 500{\rm{ kg}} \cdot {\rm{m/s}}\left( {\hat x} \right)\\\\ = 500{\rm{ kg}} \cdot {\rm{m/s}}\left( { - \hat x} \right)\\\end{array}

Therefore, the x component of the momentum of the piece A after the collision is pAx,f=500kgm/s{p_{{\rm{Ax,f}}}} = - 500{\rm{ kg}} \cdot {\rm{m/s}}.

Ans:

The component of the momentum pA,f{p_{{\rm{A,f}}}} of piece A after the explosion is 500kgm/s - 500\,{\rm{kg}} \cdot {\rm{m/s}}.

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