# Initial velocities of two particles

Two particles, of mass 1 kg and 3 kg, are traveling towards each other with velocities of 100 m/s and -40 m/s respectively. Using the laws of physics (e.g. conservation of energy & momentum) calculate the following:

a) What are the initial energies of the particles?

b) What are the final velocities of the particles?

c) Which particle gained energy, and which lost energy?

#### Solution Preview

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The initial momentum is given by:

Pi = m1*v1 + m2*v2

= (1 kg)*(100 m/s) + (3 kg)*(-40 m/s)

= 100 kg*m/s - 120 kg*m/s

= - 20 kg*m/s

The initial energy is given by:

Ei = 0.5*m1*v12 + 0.5*m2*(v22)

= 0.5*(1 kg)*(100 m/s)2 + 0.5*(3 kg)*(-40 m/s)2

= (0.5 kg)*(10,000 m2/s2) + (1.5 kg)*(1600 m2/s2)

= 5000 J + 2400 J

= 7400 J

This is also equal to the sum of the initial energies of the particles:

Ei = Ei,1 + Ei,2 = 5000 J + 2400 J = 7400 J

We can now plug the initial value of momentum into our momentum conservation equation:

Pf = m1*v1 + m2*v2 = -20 kg*m/s

(1 kg)*v1 + (3 kg)*v2 = -20 kg*m/s

To simplify, let's divide this equation by (1 kg):

v1 + 3*v2 = -20 m/s

We can also solve this equation for v1

v1 = -3*v2 - 20 m/s

Now let's plug the initial value of energy into our energy conservation equation:

Ef = 0.5*m1*v12 + 0.5*m2*v22 = 7400 J = 7400 kg*m2/s2

0.5*(1 kg)*v12 + 0.5*(3 kg)*v22 = 7400 kg*m2/s2

To simplify this equation, let's divide by 0.5*(1 kg) = 0.5 kg

v12 +3*v22 = 14800 m2/s2

Now let's plug ...