# How would you include gravity in a momentum problem?

Say you have a big ball of mass $m_1$ and a little ball on top of that of mass $m_2$ (assume they are a small distance apart, like $1~\mathrm{mm}$). Now lets drop these from a height of $h$ so that the big ball will bounce off the ground and collide into the little ball in an elastic collision.

Now I know gravity would play a key role in this example but how would one perform calculations with it? I know $F=p/t$ and momentum will not be conserved since there is an external force (gravity). So, knowing this how can one determine the height each ball will rise after the collision?

Since both balls suffer same acceleration due to gravity, their relative acceleration is zero. This means that to analyse relative motion of balls you may pretend that there is no gravity. This is equivalent to switching to a free-fall frame, which incidentally is not an inertial frame, but that does not matter so far as you want to solve kinematic problems.

• I couldn't help but to see the answers to this question. It relates to what you and I spoke of with the cylinders except gravity was not the thing canceled out in my case. Sep 26 '16 at 4:44

Use kinematics to find the velocity of both balls, so that tells you their momentum, except the lower one has its momentum reversed when it bounces. Then conserve that total momentum, and kinetic energy, during the collision at the ground. Then solve the kinematics of what follows. (By kinematics, I mean the fomulae that involve a known acceleration of g, like v^2 = 2gh.) In short, I think the key answer to your question is that conservation of energy is global and can be used at any location, but conservation of momentum is only useful in the instant of a collision.

Since the collision is elastic, you have 2 equations right off the bat. First, we have the conservation of kinetic energy at the moment where the bigger ball colliding with the lighter ball as the heavier ball moves up (dropped factor of 1/2): $$m u_{1}^{2} + M u_{2}^{2} = m v_{1}^{2} + M v_{2}^{2}$$ where $u$ is velocity before collision and $v$ is velocity after collsion, $m$ and $M$ are the small ball and big ball respectively. Next, we have the conservation of momentum at this moment as well: $$m u_{1}+ M u_{2} = m v_{1} + M v_{2}$$ The arrangement of the 2 equations into $v_{1}$ and $v_{2}$ will be left as an exercise for the reader (always wanted to type that!) and as a hint, $$v_{1}=\frac{m-M}{m+M}u_{2}+\frac{2M}{m+M}u_{2}$$ Note that I have not included gravity yet. To include the contribution from gravity, include it as an gravitational potential energy term: $mgh=\frac{1}{2}mv^{2}$

Assuming the radius of both balls is $\ll h$ then they reach the ground with approx. the same speed $u$ given by $u^2=2gh$. If $m_1$ collides elastically with the ground it rebounds with speed $-u$. There is then a collision between $m_1$ and $m_2$. The relative velocity of approach is $2u$. Suppose that after the collision $m_1$ and $m_2$ move upwards with speeds $v_1$ and $v_2$. The relative velocity of separation is then $v_2-v_1$. Because the collision is elastic, the coefficient of restitution is 1, so
$v_2-v_1=2u$.
Momentum is conserved so
$m_1u-m_2u=m_2v_2+m_1v_1$.

Substitute for $v_1$ or $v_2$ then you can find each. Use $v^2=2gh$ to find the height reached by each ball.