A 50 gram bullet is fired into a block of mass 4.0kg, initially at rest at the e

Question

A 50 gram bullet is fired into a block of mass 4.0kg, initially at rest at the edge of a frictionless table of height 0.8m. The bullet remains in the block. and after impact the block lands a distance of 1.4 m from the bottom of the table. determine the loss of kinetic energy of the system, and using the conservabito of energy for the bullet-block system. determine the speed of the system before it strikes the ground.

Explanation / Answer

let the block take t sc to fall 0.8m =>By s = ut + 1/2gt^2 =>0.8 = 0 + 1/2 x 9.8 x t^2 =>t = v0.16 =>t = 0.40 sec Let the velocity of the (mass+bullet) after collision was v m/s =>By R = {Ux} x t =>1.4 = v x 0.40 =>v = 3.46 m/s Let the velocity of the bullet was u m/s before the collision,By the law of momentum conservation:- =>mu = (M+m)v =>50 x 10^-3 x u = (4 + 50 x 10^-3) x 3.46 =>u = 280.26 m/s [Ans A] (B) KE(initial) = 1/2mu^2 = 1/2 x 50 x 10^-3 x (280.26)^2 = 1963.64 J & KE(final) = 1/2(M+m)v^2 = 1/2( 4 + 50 x 10^-3) x (3.46)^2 = 24.24 J =>KE(loss) = KE(initial) - KE(final) = 1939.40 J (C) By the law of energy conservation:- =>KE(final) + PE(final) = KE(initial) + PE(initial) =>1/2mv^2 + 0 = 1/2mu^2 + mgh =>v^2 = u^2 + 2gh =>v^2 = (3.46)^2 + 2 x 9.8 x 0.8 =>v = v27.65 =>v = 5.26 m/s

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