s shown in the fgure below, e builet is fired at and passes through a piece of t

Question

s shown in the fgure below, e builet is fired at and passes through a piece of target paper suspendes ty a massess strng. The blet as a mass m,a spees v tefore the celision with the target, and a speed0.536/v·fter passing though the target. The conson is neemt,d dng the celser, the amount of energy lost is eqial to s fraction [10.49 l of the kinetic energy of the bulet betore the cofision Oetermine the mass M o the target and the sceed V of the target the insta after the Mary sides dowm a snow-covered lon a large pece of cardboand and then sides across a trseen pond at a constant velocty of-27 m/s fter Mary hes acted the botam of the Nt and is widing acrois the ce, Sue runs after her at a velooity of+47 m's and hops on the cendboard.Now fet do the two of them ide across the ice bogther on the cardoard? Mary's mass is 70 kp and Sue's is 52 kg. ignore the mass of the cardboard and ny friction between the cardboard and the snow and/or loe. ndicate he direction with the sign of your noer enswers to at st 3 decimals) A tNin blbck of soft wood with a mass of 0 070 kp rests on a hariontal frictioniesa surface. A bulet wt The speed of the biack 20 m's mmediahely after he balet eits the bck a) Onermne the speed or the bulet as E eaits the bsck. greater than the initiel kne ne end an ietiel veoty of s.40 s while the second plever hes a12 kg mass and nl eoty of-2 m/s What is ther velocity Just after impect they ting tagetterde the directian with the ss f

Explanation / Answer

7.

M = mass of block = 0.070 kg

Vi = initial velocity of block = 0 m/s

Vf = final velocity of block = 20 m/s

m = mass of bullet = 4.67 x 10-3 kg

vi = initial velocity of bullet = 613

vf = final velocity of bullet = ?

using conservation of momentum

m vi + M Vi = m vf + M Vf

(4.67 x 10-3) (613) + (0.070) (0) = (4.67 x 10-3) vf + (0.070) (20)

vf = 313.2 m/s

b)

less than the initial kinetic energy

because some kinetic energy is lost during the collision

c)

KEi = (0.5) (m v2i + M V2i ) = (0.5) ((4.67 x 10-3) (613)2 + (0.070) (0)2 ) = 877.42 J

KEf = (0.5) (m v2f + M V2f ) = (0.5) ((4.67 x 10-3) (313.2)2 + (0.070) (20)2 ) = 243.1 J

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