A vessel at rest explodes, breaking into three pieces. Two pieces, having equal

Question

A vessel at rest explodes, breaking into three pieces. Two pieces, having equal mass, fly off perpendicular to one another with the same speed of 80 m/s. The third piece has three times the mass of each other piece. What is the magnitude of its velocity immediately after the explosion? The correct velocity is 37.71236 m/s. What is its direction at this time? __________________degrees (measured as a positive angle from the direction of either of the smaller pieces) I tried 45 degrees for the angle, but it is not correct...

Explanation / Answer

Leave the formulae aside for a moment and try to intuit the problem. Then the formulae should make more sense. The explosion is gonig to be symmetric. Since it started with 0 momentum, the sum of the pieces will also have to be 0 momentum. i.e. everything will have to cancel out. The vector sum of the known pieces m(a) and m(b) has to be equal and opposite to the unknown third piece's m(c) vector to maintain 0 momentum. This means it has to be opposite in both angle AND magnitude. Code: O m(c) m(a) / 90 / .---* | /| | / | 90 |/ | ---. m(b) / = sum of vectors of known m(a) and m(b) / = vector of unknown m(c) angle of / must be equal & opposite angle of / magnitude of / must be equal to magnitude of / times the ratio of masses Momentum comes in when we calculate the magnitude of the unknown vector. m(c)'s momentum will have to exactly balance the sum of the momenta of the m(a) and m(b). So m(c)v(c) = m(a)v(a)+m(b)v(b). Tripling the object's mass while maintaining the same momentum will require the object to go one third as far.

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