Question

Problem 14.15 A 13-kg projectile is passing through the origin O with a velocity v0 = (35 m/s)i when it explodes into two fragments A and B, of mass 5 kg and 8 kg, respectively. Knowing that 3 s later the position of fragment A is (90 m, 7 m, –14 m), determine the position of fragment B at the same instant. Assume 9.81m/s y a g   2 and neglect air resistance

Answer 1

Answer:

(114.38, -4.37, -63)m

Explanation:

Let $v_{xA}, v_{yA}, v_{zA}, v_{xB}, v_{yB}, v_{zB}$ be the velocities of object A and B at the instant right after the explosion, respectively. If we ignore air resistance and take gravity in the z direction into account then the equation of motion for object A can be written as

$s_A = <x_A , y_A, z_A> = <v_{x_A}t, v_{yA}t, v_{zA}t - gt^2/2> = <90, 7, -14>m$

$v_{xA}t = 90 \rightarrow v_{xA} = 90 / t = 90 / 3 = 30m/s$

$v_{yA}t = 7 \rightarrow v_{xA} = 7 / t = 7 / 3 = 2.33m/s$

$v_{zA}t - gt^2/2 = -14$

$v_{zA}3 - 9.81*3^2/2 = -14$

$3v_{zA} = 9.81*3^2/2 - 14 = 30.15$

$v_{zA} = 30.15 / 3 = 10.05 m/s$

We can apply the law of momentum conservation during the explosion to solve for the initial velocity of B

$M\vec{v} = m_A\vect{v_A} + m_B\vec{v_B}$

From here we can derive the 3 equation at 3 directions

1. In the x direction:

$Mv_x = m_Av_{xA} + m_Bv_{xB}$

$13*35 = 5*30 + 8v_{xB}$

$8v_{xB} = (13*35)-(5*30) = 305$

$v_{xB} = 305 / 8 = 38.13 m/s$

So after 3s object B would be at 38.13*3 = 114.38 m in the x-direction

2. In the y direction:

$Mv_y = m_Av_{yA} + m_Bv_{yB}$

$13*0 = 5*2.33 + 8v_{xB}$

$8v_{xB} = -(5*2.33) = -11.65$

$v_{xB} = -11.65 / 8 = -1.456 m/s$

So after 3s object B would be at -1.456*3 = -4.37 m in the y direction

3. In the z direction:

$Mv_z = m_Av_{zA} + m_Bv_{zB}$

$13*0 = 5*10.05 + 8v_{xB}$

$8v_{xB} = -(5*10.05) = -50.25$

$v_{xB} = -50.25 / 8 = -6.28 m/s$

So after 3 s object B would be at $(-6.28)*3 - 9.81*3^2/2 = -63 m$

Answer 2

Answer:

Position of fragment B after 3 seconds will be (114.375m,-76.11m,8.75m)

Explanation:

Motion of mass center:

It moves as if projectile has not exploded

r = v0 t i - 1/2 g t^2 j

r = (35m/s)*(3s) i - 1/2*(9.81m/s^2)*(3s^2) j

r = (105m) i - (44.145m) j

According to equation:

mr = ∑mi ri

where m is mass

mr  = mArA + mBrB

13(105i-44.145j) = 5(90i+7j-14k) + 8rB

1365i - 573.885j - 450i - 35j +  70k = 8rB

8rB = 915i - 608.885j + 70k

rB = (114.375m)i - (76.11m)j + (8.75m)k