6 CONSERVATION OF MOMENTUM 6.7 Collisions in 2 - dimensions

i (^2)
After the skater catches the ball, the combined momentum of the skater and the
ball is
p 3 = p 1 + p 2 = (960, 60) N s.
This follows from momentum conservation. The final speed of the skater (and
the ball) is (^)
vJ =
|p 3 |

√
9602

• 602
  = 6.87 m/s.
  M + m 120 + 20
  The final direction of motion of the skater is parameterized by the angle θ (see
  the above diagram), where
  θ = tan−^1
  |p 2 |
  = tan−^1
  60
  !
  = 3.58◦.
  |p 1 | (^960)
  Worked example 6.4: Bullet and block
  Question: A bullet of mass m = 12 g strikes a stationary wooden block of mass
  M = 5.2 kg standing on a frictionless surface. The block, with the bullet embed-
  ded in it, acquires a velocity of v = 1.7 m/s. What was the velocity of the bullet
  before it struck the block? What fraction of the bullet’s initial kinetic energy is
  lost (i.e., dissipated) due to the collision with the block?
  Answer: Let u be the initial velocity of the bullet. Momentum conservation re-
  quires the total horizontal momentum of the system to be the same before and
  after the bullet strikes the block. The initial momentum of the system is simply
  m u, since the block is initially at rest. The final momentum is (M + m) v, since
  both the block and the bullet end up moving with velocity v. Hence,
  m u = (M + m) v,
  giving
  u =
  M + m
  v =
  (0.012 + 5.2) × 1.7
  = 738.4 m/s.
  m 0.012
  The initial kinetic energy of the bullet is
  K =
  1
  m u^2 = 0.5 × 0.012 × 738.4^2 = 3.2714 kJ.