A bullet of mass 10 gm strikes a target a velocity of 400 m-s and loses half of its initial velocity- The loss of kinetic energy in joule is-800400200600

Question

Toppr Admin · Accepted Answer

-x394-K-E-12m-V2final-x2212-V2initial-12-xD7-10-x2212-2-4002-2-x2212-400-2-12-xD7-10-x2212-2-xD7-16-xD7-104-xD7-x2212-34-x2212-600J-x21D2- Loss -600J

A bullet of mass $10 g m$ strikes a target at a velocity of $400 m / s$ and loses half of its initial velocity. The loss of kinetic energy in joule is:
$800$
$400$
$200$
$600$

$Δ K . E = \frac{1}{2} m (V_{f i n a l}^{2} - V_{i n i t i a l}^{2})$

$= \frac{1}{2} \times 10^{- 2} ((\frac{400}{2})^{2} - (400)^{2})$

$= \frac{1}{2} \times 10^{- 2} \times 16 \times 10^{4} \times \frac{- 3}{4}$

$= - 600 J$

$\Rightarrow$ Loss $= 600 J$

A bullet of mass 10gm strikes a target at a velocity of 400m/s and loses half of its initial velocity. The loss of kinetic energy in joule is:800400200600

ΔK.E=12m(V2final−V2initial)=12×10−2((4002)2−(400)2)=12×10−2×16×104×−34=−600J⇒ Loss =600J

A bullet of mass $10 g m$ strikes a target at a velocity of $400 m / s$ and loses half of its initial velocity. The loss of kinetic energy in joule is:
$800$
$400$
$200$
$600$

$Δ K . E = \frac{1}{2} m (V_{f i n a l}^{2} - V_{i n i t i a l}^{2})$

$= \frac{1}{2} \times 10^{- 2} ((\frac{400}{2})^{2} - (400)^{2})$

$= \frac{1}{2} \times 10^{- 2} \times 16 \times 10^{4} \times \frac{- 3}{4}$

$= - 600 J$

$\Rightarrow$ Loss $= 600 J$