Find the kinetic energy - K-1-2mv2-1-2mv-v- - of a particle of mass 200 g moving with velocity 3-5-4k- m-s-

The kinetic energy of an object of mass, m moving with a velocity of 5 m s⁻¹ is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?

Q. A particle of mass

100 g

moving with a speed

1 m/s

collides perfectly inelasticaly with another particle of mass

200 g

initially at rest. Find loss in kinetic energy of the system during collision.

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Find the kinetic energy [K=12mv2=12m(→v⋅→v)] of a particle of mass 200g moving with velocity 3^i−5^j+4^km/s.

The correct option is B 5JSpeed =∣→v∣=√9+25+16=5√2m/sK.E.=12mv2=12×200×10−3×50J=5J

Find the kinetic energy $[\begin{matrix} K = \frac{1}{2} m v^{2} = \frac{1}{2} m (\to v \cdot \to v) \end{matrix}]$ of a particle of mass $200 g$ moving with velocity $3^i - 5^j + 4^k m / s$ .

The correct option is B $5 J$
Speed $=∣ \to v ∣= \sqrt{9 + 25 + 16} = 5 \sqrt{2} m / s$
$K . E . = \frac{1}{2} m v^{2} = \frac{1}{2} \times 200 \times 10^{- 3} \times 50 J = 5 J$