A bullet of mass 10   g pierces through a plate A of mass 500   g and then gets embedded into a second plate B of mass 1 . 49   kg as shown in the figure. Initially the two plates A and B are at rest and move with same velocity after collision. The percentage loss in the initial…

Question

A bullet of mass 10 &#160; g pierces through a plate A of mass 500 &#160; g and then gets embedded into a second plate B of mass 1 . 49 &#160; kg as shown in the figure. Initially the two plates A and B are at rest and move with same velocity after collision. The percentage loss in the initial kinetic energy of the bullet when it is between the plates A and B is_______. (Neglect any loss of material of the plates during the collision)

Aicharya · Accepted Answer

Let us suppose that the mass of the bullet is, $m = 10 g = 10^{- 2} kg$ and $M_{A} = \frac{1}{2} kg, M_{B} = 1.49 kg$ ,

now the loss of kinetic energy of a bullet is,

$∆ K . E . \times 100 = - (\frac{\frac{1}{2} m {v_{1}}^{2} - \frac{1}{2} m u^{2}}{\frac{1}{2} m u^{2}} \times 100) = - (\frac{{v_{1}}^{2} - u^{2}}{u^{2}} \times 100) ∆ K . E . \times 100 = - (\frac{{v_{1}}^{2}}{u^{2}} - 1) \times 100$

for plate $A$ conservation of momentum,

$m u = m v_{1} + M_{A} v \Rightarrow m (u - v_{1}) = M_{A} v$

for the plate, $B$ the bullet is combined with plate,

$m v_{1} = (m + M_{B}) v$

now from above two-equation of plates,

$\frac{u - v_{1}}{v_{1}} = \frac{M_{A}}{m + M_{B}} = \frac{0.5}{0.01 + 1.49} = \frac{0.5}{1.50} = \frac{1}{3}$

$3 u - 3 v_{1} = v_{1} \Rightarrow v_{1} = \frac{3}{4} u$

$∆ K . E . \times 100 = - (\frac{{(\frac{3}{4} u)}^{2}}{u^{2}} - 1) \times 100 = 43.75 %$

A bullet of mass 10   g pierces through a plate A of mass 500   g and then gets embedded into a…

Solution

A bullet of mass 10 &#160; g pierces through a plate A of mass 500 &#160; g and then gets embedded into a…

Solution

A bullet of mass 10 g pierces through a plate A of mass 500 g and then gets embedded into a…