Let speed of bullet-block combination is vi and bullet intial speed is vb
First find the deceleration of the block after the collision , consider that the only force affecting the motion is the frictional force , so :
F = µk (mg) = 0.75(mg) = ma
a = 0.75g = 0.75 (9.8) = 7.35 m/sec2
now using netwon’s equation and that deceleration to find speed v of bullet-block just after the collision :
vf2 = vi2 + 2ax
0 = vi2 – 2(7.35) (6.5)
hence vi = 9.77 m/sec
now we should apply the conservation of linear momentum to obtain vb at the moment of collision :
Momentum before collision = Momentum after collision
mbvb + MwVw = (mb + Mw) vi (Mw is block mass and mb is bullet mass)
0.0162 vb + 0 = (0.0162 + 0.716) (9.77)
hence vb = 441.57 m/s
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