Question

7. X You decide to try out your new rifle on a test range. The manufacturer says that a mo=4 g bullet is fired initially at a velocity of Voi=400 m/s. You decide to use this information to find the coefficient of friction between a block of wood and a concrete patio on which the block, m=0.8 kg, rests. The bullet, when fired, passes entirely through the block and emerges from the other side at 120 m/s. The block of wood, initially at rest, slides 0.45 m before stopping. A) Find the speed of the block just after the buHet passes through. B) Find an expression for the coefficient of kinetic friction and compute its value C) How much kinetic energy does the bullet lose? D) What is the kinetic energy of the block just after the bullet exits the block?

Answer 1

Giren mo= 4g = 0.004kg You = 40tups mwoording TAV ? Vw, ao $=0.45, Partifa) Apply conservation of momentum modo, t may w;= move to how to (0.004 kg/foom/s)+ 0 = (ovoo fry )x120m/s + 0.8 X Viwa Voy 0.004x400--0004X120 Par (B for block - Change inn Varg lumis, speed of block just atter billetleaves it for block to momentum 4-., V.1.4 mis AP= P.P = 0.8[( 145-607 $= 0.45m We knew 50.8(196) V²4²zzas AP= 1.56 |(1.4 m/s) - 10,5 = 2 xa x0.45 For exerted on bleek (Net force) a = (64) .jame For the boot = 2,450 Werte done by tris force on block is of block also F.s. = I mutta ungs time taben to cover Bitsm is Veurat to V4 14-0 ME FS+mWz 12.45N)60.45)-1108) (14) It=0.64 sec - 0 018x9,81 :: 0-318_0,04 2xou? = 2.18 mise this is the acceleratio Q2. javo 7.8 MK = 0.047

Part() for bullet Conservation of energy ovou leg [6400 mi? (1204193) = 0.002 boy (145600) IDR = 291.2.I} so bultet lose this much kita part de Kinatic energy of block just after bullet exit block is given by using block final relocity from a part (a) K.E of block = Įmw (wi?? = (10.8 kg ) (1.4 m/s)? KE=0.7845 /