Physics 12 Assignment KEY
2-D Collisions & Projectile Motion
2-D Collisions:
1. Define the following terms:
 Elastic collision - a collision in which both momentum and kinetic energy are conserved
 Inelastic collision - a collision in which only momentum is conserved
 Law of Conservation of Momentum – the total momentum of all objects before a
collision equal the total momentum of all objects after the collision
 Oblique (or glancing) collision – a collision in two dimensions
2. What are the conditions for an oblique collision to be perfectly elastic?




Collision must be oblique
The two objects have identical masses
One of the masses is initially stationary
The separation angle is a right angle
3. A 25 kg object moving at 3.2 m/s east collides and sticks to an 18 kg object moving at
4.1 m/s north. What is the speed in m/s of the combined mass just after the collision? Include an
appropriately scaled momentum vector diagram in your response.
4. A radioactive nucleus at rest decays into a second nucleus, an electron and neutrino. The electron
and neutrino are emitted at right angles and have momenta of 9.3 x 10 -23 kg•m/s and 5.4 x 10-23 kg
•m/s respectively. What is the magnitude and direction of the recoiling nucleus? Include a proper
momentum vector addition diagram in your answer.
Because the initial momentum is zero, the momenta of
the three products of the decay must add to zero. If we
draw the vector diagram (see right), we see that:
We find the angle from:
5. A billiard ball of mass mA = 0.400 kg moving with a speed vA = 1.80 m/s strikes a second ball,
initially at rest, mass of mB = 0.500 kg. As a result of the collision, the first ball is deflected off at an
angle of 30.0o with a speed of v’A = 1.10 m/s. (a) Taking the x-axis to be the original direction of
motion of ball a, write down the equations expressing the conservation of momentum for the
components in the x and y directions separately. (b) Solve these equations for the speed, v’ B, and
angle, θ’, of ball B. Do not assume the collision is elastic. Include a proper momentum vector addition
diagram in your answer.
6. Car A with a mass of 1.70 x 103 kg is traveling directly northeast at a speed of 14.0 m/s, and
collides with car B with a mass of 1.30 x 103 kg that is traveling directly south at a speed of 18.0 m/s.
The two cars stick together during the collision. With what velocity does the tangled mass of metal
move immediately after the collision? Include a proper momentum vector addition diagram in your
answer.
Vector addition diagram:
pAx
pA
pAy
pB
45o
pty

pt
ptx
p A  mAv A
p B  mB v B
p A  (1700kg)(14.0m / s)
p B  (1300kg)(18.0m / s)
p A  23800kgm / s
p B  23400kgm / s
p Ay  p Ax  23800 sin 45o  16800kgm / s
pty  p Ay  p By
ptx  p Ax  p Bx
pty  16800  23400
ptx  16800  0
pty  6600kgm / s
ptx  16800kgm / s
pt  (6600kgm / s ) 2  (16800kgm / s) 2
pt  18000kgm / s
16800
6600
  69
tan  
p  mv
18000kgm / s  (3000kg)v
v  6.0m / s
6.0m/s, [S69oE]
Projectile Motion:
7. Define the following terms:



Trajectory - the path described by an object moving due to a force or forces
Projectile - an object that is given an initial thrust and allowed to move through space
under the force of gravity only
Range - the horizontal distance a projectile travels
8. A ball is thrown horizontally from the roof of a building 56 m tall and lands 45 m from the base.
What was the ball’s initial speed?
We choose a coordinate system with the origin at the release point, with x horizontal and y
vertical, with the positive direction down.
We find the time of fall from the vertical displacement:
1
2
1
2
The horizontal motion will have constant velocity. We find the initial speed from
9. A 0.52-kg ball is kicked on a level field with a speed of 12 m/s at an angle of 63° above the ground.
How long, in seconds, will the ball be in the air during its flight?
Since the ball is kicked from ground level and lands at ground level, one can use the
symmetry of its parabolic trajectory. The initial vertical velocity is:
vyi = (12 m/s)(sin 63o) = 10.69 m/s
At the top of its trajectory, then:
vyf = 0 m/s
Thus, ∆t = (vyf – viy)/g = (0 m/s – 10.69 m/s)/(-9.81 m/s2)
∆t = 1.0899 s.
Due to symmetry, ∆t = 1.0899 s x 2 = 2.18 s
10. A projectile is shot from the edge of a cliff 125 m above
the ground level with an initial speed of 105 m/s at an
angle of 37.0o with the horizontal, as shown in the figure
below.
(a) Determine the time taken by the projectile to hit point P
at ground level.
(b) Determine the range X of the projectile as measured
from the base of the cliff.
At the instant just before the projectile hits point P, find (c)
the horizontal and the vertical components of its velocity,
(d) The magnitude of the velocity, and
(e) the angle made by the velocity vector with the
horizontal.
(a) We choose a coordinate system with the origin at the base of the cliff, with x horizontal and
y vertical, with the positive direction up. We find the time required for the fall from the vertical
motion:
which gives t = – 1.74, 14.6 s.
Because the projectile starts at t = 0, we have t = 14.6 s .
(b) We find the range from the horizontal motion:
(c) For the velocity components, we have
(d) When w e combine these components, we get
(e) We find the angle from