Question

A body of mass 2g, moving along the positive x-axis in gravity free space with velocity 20cms^-1 explodes at x=1m, t=0 into two pieces of masses 2//3g and 4/3g. After 5s, the lighter piece is at the point (3m, 2m, -4m). Then the position of the heavier piece at this moment, in metres is

Answer 1

Answer:

Position of heavy mass after t = 5 s is given as

$r = (\frac{3}{2} , -1, 2)$

Explanation:

Here we know that body explodes due to its internal forces

so there is no external force on it

and hence its COM will always move with same speed in same direction

so position of center of mass after t = 5 s is given as

$\Delta x = 20(5) = 100 cm$

$x - 1 = 1$

x = 2

now we have

$r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}$

$2 m = \frac{(2/3)(3, 2, -4) + (4/3)(x, y, z)}{2}$

so we have

$(4, 0, 0) = \frac{2}{3}(3, 2, -4) + \frac{4}{3}(x, y, z)$

$4 = 2 + \frac{4x}{3}$

$x = \frac{3}{2}$

$0 = \frac{4}{3} + \frac{4}{3}y$

$y = -1$

$0 = -\frac{8}{3} + \frac{4}{3} y$

$y = 2$

so position of heavy part is given as

$r = (\frac{3}{2} , -1, 2)$

#Learn

Topic : Center of mass

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Answer 2

Answer:

r=3/2

y=2

Explanation:

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