A 5,20 g bullet moving at 672 m/s strikes a 700 g wooden block at rest on a fric

Question

A 5,20 g bullet moving at 672 m/s strikes a 700 g wooden block at rest on a frictionless surface. The bullet emerges traveling in the same direction with its speed reduced to 428 m/s.

a) What is the resulting speed of the block?

b) What is the speed of the bullet-block center of mass?

Explanation / Answer

from conservation of momentum; initial momentum of bullet +initial momentum of block = final momentum of bullet+final momentum of block => m1*vi1 + m2*vi2 = m1*vf1 + m2*vf2 vi1 = 672 m/s vi2 = 0 vf1 = 428 m/s vf2 = v => 5.2*10^-3*672 +0.7*0= 5.2*10^-3*428 + 0.7*v => v = 1.813 m/s velocity of block = 1.813 m/s b). speed of centre of mass = [(mass of bullet*speed of bullet)+(mass of block*speed of block)]/total mass of bullet & block = [(m1*vf1)+(m2*vf2)]/(m1+m2) = (0.7*1.813)+(5.2*10^-3*428)/(5.2*10^-3+0.7) = 4.956 m/s

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