A Classical Approach of Newtonian Mechanics

6 CONSERVATION OF MOMENTUM 6.7 Collisions in 2 - dimensions

2

×

f

The final kinetic energy of the system is

K =

1

(M + m) v^2 = 0.5 × (0.012 + 5.2) × 1.7^2 = 7.53 J.

Hence,^ the^ fraction^ of^ the^ initial^ kinetic^ energy^ which^ is^ dissipated^ is^

f =

Ki − Kf

Ki

3.2714 103 − 7.53

=

3.2714 × 103

= 0.9977.

Worked example 6.5: Elastic collision

Question: An object of mass m 1 = 2 kg, moving with velocity vi1 = 12 m/s, col-

lides head-on with a stationary object whose mass is m 2 = 6 kg. Given that the

collision is elastic, what are the final velocities of the two objects. Neglect friction.

Answer: Momentum conservation yields

m 1 vi1 = m 1 vf1 + m 2 vf2,

where vf1 and vf2 are the final velocities of the first and second objects, respec-
tively. Since the collision is elastic, the total kinetic energy must be the same
before and after the collision. Hence,
1
m^ v^2 =^

1

m v 2 +

1

m^ v^2.^

2

(^1) i1
2
(^1) f1
2
(^2) f2
Let x = vf1/vi1 and y = vf2/vi1. Noting that m 2 /m 1 = 3 , the above two
equations reduce to
and
1 = x + 3 y,
1 = x^2 + 3 y^2.
Eliminating x between the previous two expressions, we obtain
1 = (1 − 3 y)^2 + 3 y^2 ,
or
6 y (2 y − 1) = 0,