A ball of mass 4 kg is moving in 3 dimensional plane and velocity of ball w-r-t an observer is V 1-3 -4 -5 k- and velocity of observer w-r-t ground is given as V 2-2 -2 -4 k- Find out kinetic energy of ball w-r-tg r o u n d - A- 100 JB- 38 JC -76 JD- 19 J

Given: Velocity of ball w.r.t observer, V_bo=V_1= 3î +4ĵ +5k̂ Velocity of observer w.r.t ground, V_og=V_2=2î +2ĵ 4k̂ From concept of relative motion, V_bo=V_b ...

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Standard XII

Physics

Frame of Reference

A ball of mas...

Question

A ball of mass $4 kg$ is moving in 3 dimensional plane and velocity of ball w.r.t an observer is ${\to V}_{1} = - 3^i + 4^j + 5^k$ and velocity of observer w.r.t ground is given as ${\to V}_{2} = 2^i + 2^j - 4^k$ . Find out kinetic energy of ball w.r.t ground.

100 J

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38 J

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76 J

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19 J

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Solution

The correct option is C $76 J$
Velocity of ball w.r.t ground $= V_{g}$
${\to V}_{g} = {\to V}_{1} + {\to V}_{2}$
${\to V}_{g} = (- 3^i + 4^j + 5^k) + (2^i + 2^j - 4^k)$
${\to V}_{g} = -^i + 6^j +^k$
Magnitude of ${\to V}_{g}$
$| {\to V}_{g} | = \sqrt{(- 1)^{2} + 6^{2} + 1^{2}} = \sqrt{38} m/s$
Kinetic energy of the ball
$= \frac{1}{2} m | {\to V}_{g} |^{2}$
$= \frac{1}{2} \times 4 \times (\sqrt{38})^{2}$
$= 2 \times 38$
$= 76 J$

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A ball of mass 4 kg is moving in 3 dimensional plane and velocity of ball w.r.t an observer is →V1=−3^i+4^j+5^k and velocity of observer w.r.t ground is given as →V2=2^i+2^j−4^k. Find out kinetic energy of ball w.r.t ground.

The correct option is C 76 JVelocity of ball w.r.t ground =Vg →Vg=→V1+→V2 →Vg=(−3^i+4^j+5^k)+(2^i+2^j−4^k) →Vg=−^i+6^j+^k Magnitude of →Vg |→Vg|=√(−1)2+62+12=√38 m/s Kinetic energy of the ball =12m|→Vg|2 =12×4×(√38)2 =2×38 =76 J

A ball of mass $4 kg$ is moving in 3 dimensional plane and velocity of ball w.r.t an observer is ${\to V}_{1} = - 3^i + 4^j + 5^k$ and velocity of observer w.r.t ground is given as ${\to V}_{2} = 2^i + 2^j - 4^k$ . Find out kinetic energy of ball w.r.t ground.