physics help... detailed solution pls 4.00 g bullet is moving horizontally with

Question

physics help... detailed solution pls

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 first block (mass 1250 g) (an inelastic collision) and embeds itself in the second one (mass 1430 g), the velocity of the first block after the bullet passes through is +0.450m/s. What is the velocity of the second block after the bullet embeds itself? Find the ratio of the total kinetic energy after the collisions to that before the collisions. (b) After Collisions

Explanation / Answer

mb = mass of bullet = .004 kg

m1 = mass of block 1 = 1.25 kg

m2 = mass of block 2 = 1.43 kg

a) mb*vb1 = m1*v1 + mb*vb2

.004*335 = 1.25*.45 + .004*vb2

vb2 = 195 m/s

mb*vb2 = (mb+m2)v2

v2 = .004*195/(1.43+.004) = 0.54 m/s (Ans)

b) KE after collision = 0.54*(1.43+.004) + 1.25*.45 =0.77 + 0.56 = 1.3368 J = 1.34 J

KE before collision = 0.004*335 = 1.34 J

Ratio = 1:1

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