# How is momentum conserved when the ball collides with the floor?

Jan 25, 2016

Momentum is always conserved, irrespective of the size of colliding bodies.

#### Explanation:

In a two body system, let ${m}_{1} \mathmr{and} {m}_{2}$ be respective masses of two colliding bodies, ${\vec{v}}_{1} \left(i n i t i a l\right) , {\vec{v}}_{2} \left(i n i t i a l\right)$, ${\vec{v}}_{1} \left(f i n a l\right)$ and ${\vec{v}}_{2} \left(f i n a l\right)$ be respective velocities before and after the collision respectively.

Now momentum is defined as

$\vec{p} = m . \vec{v}$

The total initial momentum

${\vec{p}}_{b e f o r e} = {m}_{1.} {\vec{v}}_{1} \left(i n i t i a l\right) + {m}_{2.} {\vec{v}}_{2} \left(i n i t i a l\right)$

And final momentum after the collision is

${\vec{p}}_{a f t e r} = {m}_{1.} {\vec{v}}_{1} \left(f i n a l\right) + {m}_{2.} {\vec{v}}_{2} \left(f i n a l\right)$

By conservation of momentum
${\vec{p}}_{b e f o r e} = {\vec{p}}_{a f t e r}$
We obtain

${m}_{1.} {\vec{v}}_{1} \left(i n i t i a l\right) + {m}_{2.} {\vec{v}}_{2} \left(i n i t i a l\right) =$
${m}_{1.} {\vec{v}}_{1} \left(f i n a l\right) + {m}_{2.} {\vec{v}}_{2} \left(f i n a l\right)$

In the given problem, the two bodies are 1. a ball and 2. floor. As the floor is rigidly connected to the building, standing on the earth's surface it is appropriate to assume that the second colliding body is earth. Also that in a frame reference to earth it is the ball which moves and earth is at rest. The expression reduces to

${m}_{b a l l} . {\vec{v}}_{b a l l} \left(i n i t i a l\right) = {m}_{b a l l} . {\vec{v}}_{b a l l} \left(f i n a l\right) + {m}_{e a r t h} . {\vec{v}}_{e a r t h} \left(f i n a l\right)$

For simplicity and for sake of argument assuming that the ball bounces back elastically in the reverse direction after collision,

${\vec{v}}_{b a l l} \left(f i n a l\right) = - {\vec{v}}_{b a l l} \left(i n i t i a l\right)$
Inserting this in the equation we obtain

${m}_{e a r t h} . {\vec{v}}_{e a r t h} \left(f i n a l\right) = 2 {m}_{b a l l} . {\vec{v}}_{b a l l} \left(i n i t i a l\right)$
$\implies {\vec{v}}_{e a r t h} \left(f i n a l\right) = 2 {m}_{b a l l} / {m}_{e a r t h} . {\vec{v}}_{b a l l} \left(i n i t i a l\right)$

We know that mass of earth is 5.972 × 10^24 kg.

For a bowling ball for kids the ratio $2 {m}_{b a l l} / {m}_{e a r t h} \approx {10}^{- 24}$

This expression has been derived for a special case.
Generalizing, it can be seen that after the collision speed of earth is very very small quantity because of the ratio of masses of the ball an earth being present in the equation. We are not able to measure such a quantity with respect to speed of earth in its orbit around the Sun which is $\approx 3.0 \times {10}^{4} m / s$.

Having said, we have no reason to believe that the momentum is not conserved.