A 30 kg projectile moving horizontally with a velocity ${\to v}_{0} = (120 m / s)^i$ explodes into two fragments A and B of masses 12 kg and 18 kg, respectively. Taking point of explosion as origin and knowing that 3 s later the position of fragment A is (300, 24, –48) m, determine the position of fragment B at the same instant.

(360, -45, 32)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(400, -91, 32)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(400, -91, 48)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(360, -91, 48)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B (400, -91, 32)
Initial coordinates of CM = (0, 0, 0)
x coordinates of CM after 3 s,
$x_{C M} = 120 \times 3 = 360 m$
y coordinates of CM after 3 s,
$y_{C M} = - \frac{1}{2} g t^{2} = - 45 m$
After 3 s,
$(m_{1} + m_{2}) x_{C M} = m_{1} x_{1} + m_{2} x_{2}$
$x_{2} = \frac{30 \times 360 - 12 \times 300}{18} = 400 m$
Similarly, $(m_{1} + m_{2}) y_{C M} = m_{1} y_{1} + m_{2} y_{2}$
$- 30 \times 45 = 12 \times 24 + 18 \times y_{2}$
$y_{2} = - 91 m$
and in z coordinate, $m_{1} z_{1} + m_{2} z_{2} = 0 \Rightarrow z_{2} = 32 m$

Suggest Corrections

Similar questions

{\to v}_{0} = (120 m / s)^i

{\to v}_{0} = (120 m / s)^i

30 k g

48 m / s

18 k g

12 k g

45^{0}

\frac{1}{8} k g a n d \frac{7}{8} k g

Join BYJU'S Learning Program