We find the force of the first object using the Impulse-Momentum Theorem.

• The first object has a change in velocity of -3 m/s - 9 m/s = -12 m/s.
• Its change in momentum is therefore

• `dp1 = m1 `dv1 = 6 kg * -12 m/s = -72 kg m/s.

• The average force exerted on this object by the second is therefore

• Fave12 = `dp / `dt = -72 kg m/s / ( .049 sec) = -1469.388 kg m/s^2 = -1469.388 Newtons.

We find the change in momentum of the second object using the Impulse-Momentum Theorem, and use this change to find the change in velocity and subsequently the final velocity of this object.

• The average force exerted by the first object on the second will be equal and opposite to the force exerted by the second on the first:

• Fave21 = -Fave12.

• Since this force acts for exactly the same time interval as the first, the corresponding change in the momentum of the second object will be

• `dp2 = Fave12 * `dt = 1469.388 Newtons * .049 sec = 72 Newton seconds = 72 kg m / s.

• Since the second object has mass 9 kg, its change in velocity will be

• `dv2 = `dp2 / m2 = 72 kg m/s / ( 9 kg ) = 8 m/s.

• The final velocity the second object is therefore

• vf2 = v02 + `dv2 = -7 m/s + 8 m/s = 1 m/s.

We could have reasoned this problem out more simply by noting that since the forces are equal and opposite, and the time interval identical for both forces, the momentum changes `dp2 = Fave21 * `dt and `dp1 = Fave12 * `dt must be equal and opposite.

• Thus `dp2 = m2 `dv2 must equal -`dp1 = -m1 `dv1:

• m2 `dv2 = -m1 `dv1.

• It follows them that

• 9 kg * `dv2 = - 6 kg * -12 m/s

• and

• `dv2 = - ( 6 kg / 9 kg) * -12 m/s = 8 m/s.

If no energy is added to the system (as for example in a collision involving an explosion, which might typically convert chemical potential energy to kinetic) then the total kinetic energy of the system must not increase. The total kinetic energy might decrease by being partially or wholly converted to thermal energy, sound energy, etc..

• The total kinetic energy before collision is easily computed as the sum of the kinetic energies of objects before collision:

• KEbefore = .5 * 6 kg * ( 9 m/s)^2 + .5 * 9 kg * (-7 m/s)^2 = 463.5 Joules.

In general when two objects collide they exert equal and opposite forces on one another. If we let Fave12 and Fave21 represent respectively the forces exerted on object 1 by object 2 and by object 2 on object 1, then this condition is expressed as

• Fave12 = - Fave21.

If the first object has constant mass m1 and initial velocity v1, and if its after-collision velocity is v1', then its change in momentum is

• `dp1 = m1 * (v1' - v1).

If the duration of contact is `dt, then the average force is

• Fave12 = `dp1 / `dt = m1 * (v1' - v1) / `dt.

Since the force exerted on object 2 by object 1 is Fave21 = - Fave12, we have

• Fave21 = - m1 * (v1' - v1) / `dt.

If the before- and after-collision velocities of the second object are v2 and v2', we also see that `dp2 = m2 * ( v2' - v2) so

• Fave21 = m2 * (v2' - v2) / `dt.

It follows that the two expressions for Fave21 are equal:

• -m1 * (v1' - v1) / `dt = m2 * ( v2' - v2) / `dt.

This equation can be expressed as

• m2 `dv2 / `dt = m1 `dv1 / `dt.

Both sides can be multiplied by `dt to give us the equation m2 `dv2 = -m1 `dv1, which is the same as

• `dp2 = - `dp1.

We thus see that the objects have equal and opposite changes in momentum.

From the form m2 `dv2 = -m1 `dv1 we see that

• `dv2 = - (m1 / m2) * `dv1.

The original equation -m1 * (v1' - v1) / `dt = m2 * ( v2' - v2) / `dt can be simplified by multiplying both sides by `dt to get

• -m1 * (v1' - v1) = m2 * ( v2' - v2),

which can then be expanded to give

• -m1 v1' + m1 v1 = m2 v2' - m2 v2.

If we collect the before-collision quantities on the left and the after-collision quantities on the right side of this equation we obtain

• m1 v1 + m2 v2 = m1 v1' + m2 v2'.

This form is the standard expression of the Law of Conservation of Momentum for two object in a closed system.

• We could have solved the stated problem using this form, since we were given the values of all the variables m1, v1, v1', m2, v2, and v2' except v2'.
• The solution is easily found by rearranging the equation to get

• Solution to present problem:  v2' = ( m1 v1 - m1 v1' + m2 v2 ) / m2.

In a simple collision (one not involving and explosion or some other conversion of some potential energy stored in one or the other of the to objects which comprise the system) kinetic energy must either remain the same or decrease due to dissipation (e.g., through thermal energy, wave motion, etc.).

We thus have

• KE0 = .5 m1 v1^2 + .5 m2 v2^2 >= KEf = .5 m1 v1'^2 + .5 m2 v2'^2.