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?
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 ...