istory Bookmarks Tools Help MasteringPhysics: CH 9, 10, t 11 Homework - Moz http

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istory Bookmarks Tools Help MasteringPhysics: CH 9, 10, t 11 Homework - Moz https://session.masteringphysics.com/myct temView assi CH9, 10 & 11 Homework Problem 9.21 Problem 9.21 A 244-kg projectile, fired with a speed of 128 m/s at a 68.0° angle, breaks into three pieces of equal mass at the highest point of its arc (where its velocity is horizontal). Two of the fragments move with the same speed right after the explosion as the entire projectile had just before the explosion; one of these moves vertically downward and the other horizontally

Explanation / Answer

A]Horizontal component of velocity = Vx = 128 cos 68 degree = 47.95 m/s

Let the speed of third be v,

momentum conservation in x direction,

m*47.95 = m/3 * 47.95 + m/3 *v cos alpha

v cos theta = (47.95*3 - 47.95) = 95.9 m/s

momentum conservation in x direction,

m/3* v sin alpha = m/3*47.95 or v sin alpha = 47.95

v = sqrt(95.9^2+47.95^2) = 107.2 m/s answer

B] alpha = arcsin(47.95/107.2) = 26.57 degree

C] energy released = - 0.5mVx^2 + 0.5*m/3*(v^2 + Vx^2 +Vx^2)

= -0.5*244*47.95^2 +0.5*244/3*(107.2^2 + 47.95^2 + 47.95^2)

= 373833 J

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