Question

W chapter 9 Problem 9.21 Problem 9.21 A207-kg projectile, fired with a speed of 145 m/s at a 62.0 o 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

Answer 1

Since at the highest point the mass will have only horizontal velocity is, v_k = V_0 cos theta = 145 cos 62 degree = 68 m/s The mass of the each fragment is, m' = 207/3 = 69kg In horizontal direction, The linear momentum of the projectile before the collision, p_i = mv_k = (207)(68) = 14076 kg m/s The linear momentum of the fragment moves horizontally before the collision, p_1 = m'v_k cos theta = (69) (68) cos 62 degree = 2202.8 kgm/s The linear momentum of the fragment moves vertically before the collision, p_2 = 0 The linear momentum of the third fragment moves vertically before the collisiorn p_3 = m'v cos theta = 69vcos theta From the conservation of linear momentum in horizontal direction, p_i = p_1 + p_2 + p_3 14076 = 2205.8 + 0 + 69vcos thate 14076 - 2202.8 = 69 v cos theta v cos theta = 11873.2/69 v cos theta = 172m/s In vertical