A Classical Approach of Newtonian Mechanics

6 CONSERVATION OF MOMENTUM 6.7 Collisions in 2 - dimensions

m 2

m 1

vi1 m 2





m 1

vf1

which has the non-trivial solution y = 1/2. The corresponding solution for x is

x = (1 − 3 y) = −1/2.

It follows that the final velocity of the first object is

vf1 = x vi1 = −0.5 × 12 = − 6 m/s.

The minus sign indicates that this object reverses direction as a result of the

collision. Likewise, the final velocity of the second object is

vf2 = y vi1 = 0.5 × 12 = 6 m/s.

Worked example 6.6: 2 - dimensional collision

Question: Two objects slide over a frictionless horizontal surface. The first object,

mass m 1 = 5 kg, is propelled with speed vi1 = 4.5 m/s toward the second object,
mass m 2 = 2.5 kg, which is initially at rest. After the collision, both objects have

velocities which are directed θ = 30◦ on either side of the original line of motion

of the first object. What are the final speeds of the two objects? Is the collision

elastic or inelastic?

y (^) vf2
x
Answer: Let us adopt the coordinate system shown in the diagram. Conservation
of momentum along the x-axis yields
m 1 vi1 = m 1 vf1 cos θ + m 2 vf2 cos θ.
Likewise, conservation of momentum along the y-axis yields
m 1 vf1 sin θ = m 2 vf2 sin θ.