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Standard XII

Physics

Velocity of COM

Three badies ...

Question

Three badies $A, B$ and $C$ having masses $10 k g, 5 k g$ and $15 k g$ representively are projected from top of a tower with $A$ vertically upwards with $10 m / s, B$ with $20 m / s 53^{o}$ above east horizontal and $C$ horizontally southward with $15 m / s$ . Find
(a) Velocity of centre of mass of the systrem.

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Solution

Given,

${\to v}_{a} = 10^k$

${\to v}_{b} = 20 cos 53^i + 20 sin 53^k$

${\to v}_{c} = 15^j$

From conservation of momentum

Total momentum = momentum of center of mass

$m_{a} {\to v}_{a} + m_{b} {\to v}_{b} + m_{c} {\to v}_{c} = (m_{a} + m_{b} + m_{c}) {\to V}_{c m}$

$10 \times 10^k + 5 (20 cos 53^i + 20 sin 53^k) + 15 \times 15^j = (10 + 5 + 15) V_{c m}$

${\to V}_{c m} = 2^i + 7.5^j + 5.99^k$

Center of mass velocity is $2^i + 7.5^j + 5.99^k$
Velocity
$2 m s^{- 1}$ toward east
$7.5 m s^{- 1}$ toward south

$5.99 m s^{- 1}$ upward

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Three badies A,B and C having masses 10 kg,5 kg and 15 kg representively are projected from top of a tower with A vertically upwards with 10 m/s,B with 20 m/s 53o above east horizontal and C horizontally southward with 15 m/s. Find(a) Velocity of centre of mass of the systrem.