A metal bullet moving at $400 m s^{- 1}$ strikes on a tree trunk and get embedded inside it. Assuming total kinetic energy is converted to heat and $50$ % of heat is absorbed by the bullet, find the increase in its temperature.(specific heat capacity of metal $= 200 J k g^{- 1} K^{- 1}$ and bullet not melts)

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Solution

K.E of bullet $= \frac{1}{2} m v^{2} = \frac{1}{2} \times m (400)^{2}$
$= 160000 \frac{m}{2}$
$= 80000 m$
Heat absorbed by bullet $= \frac{1}{2} K . E$
$m s Δ T = 4000 m$
$Δ T = \frac{40000 m}{m \times 200} = 200^{\circ}$ C

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A metal bullet moving at 400 ms−1 strikes on a tree trunk and get embedded inside it. Assuming total kinetic energy is converted to heat and 50% of heat is absorbed by the bullet, find the increase in its temperature.(specific heat capacity of metal =200 J kg−1K−1 and bullet not melts)

K.E of bullet =12mv2=12×m(400)2=160000m2=80000mHeat absorbed by bullet =12K.EmsΔT=4000mΔT=40000mm×200=200∘C

K.E of bullet $= \frac{1}{2} m v^{2} = \frac{1}{2} \times m (400)^{2}$
$= 160000 \frac{m}{2}$
$= 80000 m$
Heat absorbed by bullet $= \frac{1}{2} K . E$
$m s Δ T = 4000 m$
$Δ T = \frac{40000 m}{m \times 200} = 200^{\circ}$ C