A 225 kg projectile, fired with a speed of 109 m/s at a 66.0 angle, breaks into

Question

A 225 kg projectile, fired with a speed of 109 m/s at a 66.0 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.

Determine the magnitude of the velocity of the third fragment immediately after the explosion.

Determine the direction of the velocity of the third fragment immediately after the explosion.

Determine the energy released in the explosion.

Explanation / Answer

as in the explosion there are no external forces so change in momentum of the total body = 0 ( gravity is an external force, however in the time of a collision being considered here the change in momentum due to it will be almost 0) total horizontal momentum before = after 225 * 109(cos66) = 75 * 109cos66 + 75v vertical momentum 0 = 75 *109cos66 - 75 u assuming u is upward and is negative, the other term in eqn 2 being downward is positive net velocity = 99.12 m/s at an angle from horizontal = 26.56degrees energy released during collision is the total kinetic energy before explosion- total kinetic energy of all parts after explosion

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