D. Momentum Part II

6.3-6.4

6.3    Conservation of Momentum

40, 42, 43, 45, 46, 50, 53

42:  Remember the momentum is conserved.  Also, after he throws the hammer, his mass becomes 59.5 kg)

46:  Resolve Px direction so that m1xv1x + m2xv2x + m3xv3x = 0. 

NOTE:  v2x = 0m/s since it isn’t moving that direction.  m3 = 1.2 kg. 

Next,  Resolve Py direction so that m1yv1y + m2yv2y + m3yv3y = 0. 

NOTE:  v1y = 0 m/s since it isn’t moving in the y direction.  Again, m3 = 1.2 kg.

50:  Momentum is conserved so whatever momentum the bullet loses, the block picks up

53:  Remember that velocity of the pendulum before it is hit is zero.  The velocity at the top of its swing (after it has been hit) is also zero (but maximum potential – mgh).  As it comes back to it’s original position, it has maximum velocity AND maximum KE !  Since momentum is conserved, the initial momentum must equal the final momentum.

 

6.4    Elastic and Inelastic Collisions:

 

54, 55, 56, 58, 62, 64, 66, 71  (note:  type-o in the book!  62 and 66 are both ELASTIC collisions!)

58:  Use equations 6.15 and 6.16 on p. 191

62:  mvo1 + mvo2 = mv1 + mv2.  Solve for v2.  Realize that vo2 will be –4.0 m/s because it is heading in the –x direction.  Once you solve for v2, use Ko = ½ mvo12 + 1/2mvo22 and find the difference between Ko and K (which is 1/2mv12 + 1/2mv22.)  Remember whenever you find difference, it is FINAL – INITIAL (K – Ko)

64:  Solve for resulting velocity in the x direction and in the y direction as follows:

X direction:  mvx + MVx = (m + M) v’x          (v’x = resulting velocity in x direction)

Y direction:  mvy + MVy = (m + M) v’y

Realize that mass “M” has zero velocity in the x direction and mass “m” has zero velocity in the y direction.

66:  Subtract vectors.  DP = Pfinal – P initial.  Realize ball #1 only has an x velocity.  Ball #2 has x and y.)

71:  a)  Voblock = Vobullet/(m+M/m) where “m” is for the bullet and “M” is the mass of the block.  PLEASE CONVERT MASSES INTO KG!

b)  Vobullet = m+M/m()

c)      KElost = M/m+M

Even Answers:

40.     a

42.     moves 0.083 m/s in the opposite direction

46. 2.0 m/s, 54o above the +x axis

50.       0.33 m/s

54.       c

56.      c

58.      v1 = +1.3 m/s;  v2 = +5.3 m/s

62.       1.1 x 102J

64.     a)  v’x = 1.0 m/s;  v’y = 3.3 m/s

      b)  73o

66.  15o, 0.57 m/s