Conservation of Momentum
“in 2-D” Solutions
1. A cue ball, traveling at 70 cm/s, strikes a stationary billiard ball a
glancing blow, as shown below. Both balls travel with the same
speed after the collision, both at angles of 20o relative to the cue
ball’s original direction. Find the speed of the billiard balls after the
collision.
20o
20o
momentum is conserved in the x - direction ()
m(70 cms )  m(0 cms )  m(v cos 20 o )  m(v cos 20 o )
70 cms  2v cos 20 o
v  37.2 cms
2. Find the angle necessary for each ball in the situation above to
travel at a speed of 50 cm/s. (You may assume the same initial velocity
of the cue ball and you may assume that both ball’s travel at the same angle
relative to the cue ball’s original direction)
momentum is conserved in the x - direction ()
m(70 cms )  m(0 cms )  m(50 cms  cos  )  m(50 cms  cos  )
70 cms  2(50 cms ) cos 
cos   0.7
  cos 1 (0.7)  45.5 o
3. Two cars collide and stick to each other. The first car, with mass
1,500 kg, was traveling at 45 m/s [S 30o E]. The second car of
mass 2000 kg was traveling at 36 m/s due North. Find the velocity
of the cars after the collision.
momentum is conserved in the x - direction ()
(1500kg)( 45 ms  sin 30)  (2000kg)(0 ms )  (3500kg)v x
33,750 ms  (3500kg)v x
v x  9.643 ms
momentum is also conserved in the y - direction ()
(1500kg)( 45 ms  cos 30)  (2000kg)(36 ms )  (3500kg)v y
13,543.285  (3500kg)v y
v x  3.870 ms
v  v x  v y  10.390 ms
 vy
  tan 
 vx
-1
m



-1 3.870 s
o

  tan 
 9.643 m   21.87

s 

Therefore, v  10.390 ms [ E 21.87 o N ] or [ N 68.13o E ]
4. In a physics classroom not so far away, a 5g piece of chalk
spontaneously explodes into 3 pieces. Sound impossible? Maybe.
But it would explain the chalk residue all over Mr. L’s clothes, face,
hands, etc. A 2 g shard travels southwards at 25 m/s. Another 2g
piece travels at 20 m/s [N 40o W]. Find the velocity (magnitude &
direction) of the last piece.
momentum is conserved in the x - direction ()
(.002kg  .002kg  .001kg)(0 ms )
 (.002kg)(0 ms )  (.002kg)( 20 ms  sin 40)  (.001kg)(v x )
0  .0257 N  s  (.001kg)(v x )
v x  25.712 ms
momentum is also conserved in the y - direction ()
(.002kg  .002kg  .001kg)(0 ms )
 (.002kg)( 25 ms )  (.002kg)( 20 ms  cos 40)  (.001kg)(v y )
0  .01936  (.001kg)(v y )
v y  19.358 ms
v  v x  v y  32.18 ms
v
  tan  y
 vx
-1
m

-1  19.358 s 
  36.975o
  tan 
m 

 25.712 s 
Therefore, v  32.18 ms [ E 37 o N ] or [ N 53o E ]
Conservation of Momentum in 2D - VECTOR PROBLEMS
1. A grenade of mass 1.20 kg is at rest on a smooth, frictionless surface when it suddenly
explodes into three pieces. A 0.50 kg piece flies off horizontally to the north at 3.0 m/s,
and a 0.30 kg piece flies off horizontally to the southwest at 4.0 m/s. What is the horizontal
speed and direction of the third piece?
2. A 2000 kg car traveling east at 24 m/s enters an icy intersection and collides with a 3600 kg
truck traveling south at 10 m/s. If they become coupled together in the collision, what is
their velocity immediately after impact?
3. A billiard ball (the cue ball) travelling north at 5 m/s strikes a stationary billiard ball (the 8ball) of equal mass a glancing blow. After the collision, the 8-ball which was at rest moves
at 3 m/s [NW]. Find the velocity of the cue ball.
x:
mv  mv  mv  mv
m(0)  m(0)  m(3cos 45)  mvx
v x  2.121 ms
y:
mv  mv  mv  mv
m(5)  m(0)  m(3sin 45)  mvy
v y  2.879 ms
v  (2.121) 2  (2.879) 2  3.576 ms
  tan 1 (2.879 /2.121)  53.62 o
v  3.576 ms [E 53.62 o N]

4. A steel ball of mass 0.50 kg, moving with a velocity 2.0 m/s (East), strikes a second ball of
mass 0.30 kg, initially at rest. The collision is a glancing one, causing the first ball to be
deflected N 60° E, with a speed of 1.50 m/s. Determine the velocity of the second ball after
the collision, giving both its speed and direction.
5. A 3000 kg space capsule is traveling in outer space with a velocity of 200 m/s (East). In an
effort to alter its course, it fires a 25.0 kg projectile North at a speed of 2000 m/s. What is
the new velocity of the space capsule?
6. A nucleus, initially at rest, decays radioactively. In the process, it emits an electron
horizontally to the east, with momentum 9.0 x 10–21 kg·m/s and a neutrino horizontally to
the south, with momentum 4.8 x 10–21 kg·m/s.
a. In what direction does the residual nucleus move?
b. What is the magnitude of its momentum?
c. If the mass of the residual nucleus is 3.6 x 10–25 kg, what is its recoil velocity?