A 4.00 g bullet is moving horizontally with a velocity of +355 m/s (moving to th

Question

A 4.00 g bullet is moving horizontally with a velocity of +355 m/s (moving to the right). The bullet is approaching two blocks resting on a horizontal frictionless surface (air resistance is negligible). The bullet passes completely through the rst block (mass 1150 g) (an inelastic collision) and embeds itself in the second one (mass 1530 g), the velocity of the rst block after the bullet passes through is +0.550m/s. (a) What is the velocity of the second block after the bullet embeds itself? (b) Find the ratio of the total kinetic energy after the collisions to that before the collisions.

Explanation / Answer

Impulse I conservation must be applied : I = mbu*Vbu = 0.004*355 = 1.42 kgm/sec Impulse I1 = mb1*Vb1 = 1.150*0.55 = 0.825 kgm/sec Impulse I2 = I-I1 = 1.42-0.825 = 0.595 kgm/sec mb2 = (1530+4.0)/1000 = 1.534 kg a) Vb2 = I2/mb2 = 0.595 /1.534 = 0.388 m/sec b) Ei = 1/2mbu*Vbu^2 = 0.5*0.004*355^2 = 252.05 joule Ef = 1/2*1.150*0.550^2 + 1/2*1.534*0.388^2 = 0.174+0.115 = 0.289 Joule K = Ef/Ei = 0.289/252.05 = 1.146*10^-3

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