Two 20 kg cannon balls are chained together and fired horizontally with a veloc

Discussion Forum : Center Of Mass

Question -

Two 20 kg cannon balls are chained together and fired horizontally with a velocity of 200 m/s from the top of a 30 m wall. The chain breaks during the flight of the cannon balls and one of them strikes the ground at t = 2 s, at a distance of 250 m from the foot of the wall, and 5 m to the right of the line of fire. Determine the position of the other cannon ball at that instant. Neglect the resistance of air.

Options:

A . (550, 20, -5)

B . (250, 20, 5)

C . (550, 10, -5)

D . (550, 20, 0)

Answer: Option A
:
A
As no external force acts in z direction, hence z-coordinate of the centre of mas of the ball should be zero. To make z-coordinate zero other ball should fall symmetrically with respect to z-axis.
Hence, z-coordinate of other ball = –5 m.
The balls do not have any external force in x-direction. Hence in x-direction the centre of mass should move with constant velocity. X-coordinate of centre of mass at

t = 2 s = 200 \times 2 = 400 m

Hence,

x_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}

400 = \frac{20 \times 250 + 20 x_{2}}{20 + 20}

x_{2} = 800 - 250 = 550 m

Height fallen by centre of mass at t = 2s,

h = \frac{1}{2} \times 10 \times (2)^{2} = 20 m

Hence, y-coordinate of centre of mass

= 30 - 20 = 10 m

Hence,

y_{C M} = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}

10 = \frac{20 \times 0 + 20 \times y_{2}}{20 + 20}

\Rightarrow y_{2} = 20 m

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