A 5 gram bullet moving with an initial speed of 400m/s is firedinto and passes t

Question

A 5 gram bullet moving with an initial speed of 400m/s is firedinto and passes through a 1 Kg block. The block, initially atrest on a frictionless, horizontal surface, is connected to aspring with k = 900 N/m. If the block moves a distance of 5cm to the left after impact, find
(a) the speed at which the bullet emerges from the block (b) the energy lost in the collision
I only understand how to find the speed of the block + bulletusing the equation m1v1 = (m1+m2)v2
(a) the speed at which the bullet emerges from the block (b) the energy lost in the collision
I only understand how to find the speed of the block + bulletusing the equation m1v1 = (m1+m2)v2

Explanation / Answer

Total max energy is kinetic initial energy of thebulet Kei Kei = 0.5mVi2 =0.5 x 0.005 (400)2= 400 J after impact Kea = 0.5(M + m)Va 2 mV = (M +m)Va Va=mV /(M +m)=0.005 x 400/( 1.000 + 0.005)= Va=1.99 m/s Kea = 0.5(M + m)Va 2=0.5( 1.000 +0.005) (1.99)2 = Kea =1.99 J and potential energy of the compressed spring Pe= 0.5kx2= 0.5 x 900 x (0.05)2= 1.125J (this is inconsistant) Finally when the bullet exits in has energy Kef=Kei - (Pe +Kea) = 397 J Kef=   400 - ( 1.125 +1.99) = Kef=  0.5mVf2   and then has a velocity Vf=(2Kef/m) Vf=(2 (Kef ) /m) Now (a) the speed at which the bullet emerges from the block Vf=(2 (Kei - Kea/m) Vf=(2 (397) /0.005) Vf= 398 m/s (b) the energy lost in the collision is Pe + Ke (a) the speed at which the bullet emerges from the block Vf=(2 (Kei - Kea/m) Vf=(2 (397) /0.005) Vf= 398 m/s (b) the energy lost in the collision is Pe + Ke energy lost = 1.125 + 1.99= 3.12 J we assume no energy wasconsumed by going trough the block.

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