A particle of mass $$200$$ g is dropped from a height of $$50$$ m and another particle of mass $$100$$ g is simultaneously projected up from the ground along the same line, with a speed of $$100$$ $$m/s$$ . The velocity of the centre of mass after $$1$$ s will be:

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Infinity Learn · Accepted Answer

After t $$=$$ $$1$$ $$sec$$, the velocity of : $$200$$ g mass will be $$v1$$ $$=$$ $$0$$ $$+$$ $$10(1)$$ $$=$$ $$10$$ $$m/s$$ (down) and that of $$100$$ g mass will be $$v2$$ $$=$$ $$100$$ $$-$$ $$10(1)$$ $$=$$ $$90$$ $$m/s(up)Velocity$$ v of centre of mass will be v $$=$$ $$200(10)$$ $$+$$ $$100(-90)200$$ $$+100$$ $$=$$ $$703m/s$$ up

Motion of centre of mass

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A particle of mass 200 g is dropped from a height of 50 m and another particle of mass 100 g is simultaneously projected up from the ground along the same line, with a speed of 100 m/s . The velocity of the centre of mass after 1 s will be:

After t = 1 sec, the velocity of : 200 g mass will be $v_{1}$ = 0 + 10(1) = 10 m/s (down) and that of 100 g mass will be
$v_{2} = 100 - 10 (1) = 90 m / s (up)$
Velocity $v$ of centre of mass will be
$v$ = $\frac{200 (10) + 100 (- 90)}{200 + 100} = \frac{70}{3} m / s up$

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Motion of centre of mass

Remember concepts with our Masterclasses.

A particle of mass 200 g is dropped from a height of 50 m and another particle of mass 100 g is simultaneously projected up from the ground along the same line, with a speed of 100 m/s . The velocity of the centre of mass after 1 s will be:

After t = 1 sec, the velocity of : 200 g mass will be v1 = 0 + 10(1) = 10 m/s (down) and that of 100 g mass will be v2 = 100 - 10(1) = 90 m/s(up)Velocity v of centre of mass will be v = 200(10) + 100(-90)200 +100 = 703m/s up

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After t = 1 sec, the velocity of : 200 g mass will be $v_{1}$ = 0 + 10(1) = 10 m/s (down) and that of 100 g mass will be
$v_{2} = 100 - 10 (1) = 90 m / s (up)$
Velocity $v$ of centre of mass will be
$v$ = $\frac{200 (10) + 100 (- 90)}{200 + 100} = \frac{70}{3} m / s up$