Momentum conservation in ball and massive wall collision

Question

I know momentums definition (p=mv) and how to calculate problems involving it, but there are bits I don't understand.

Take a ball thrown at a wall for instance with mass value (5) and velocity (5). This would mean the total momentum of ball is (25). The wall is attached to the earth so the mass is huge and the wall doesn't move (momentum of (0)). Total momentum in system is now (25). This ball hits a wall and bounces back in the opposite direction.

The ball now has a momentum of (-25) so its moving in the opposite direction from which it started. The wall doesn't 'move' due to its high mass (and undetectable gain in velocity). THe total momentum of the system still has a magnitude of (25) but there has been a change in momentum of (50) for the ball. Where did the extra momentum come from? I would think that the ball would transfer all its momentum to the wall in the collision so it would stop moving (as its its momentum drops to 0) but it actually gains momentum to go in the opposite direction.

Where is this extra momentum coming from? Is it linked to newtons 3rd? But then there is a gain of acceleration...

Answer 1

The wall gained some momentum in-fact, Notation;

$v_i$ = initial velocity of ball

$v_f$ = final velocity of ball

$V$ = Final velocity of wall $$ P_{initial} = P_{final}$$ $$ mv_i = -mv_f + m_{wall} V $$

now, $ v_i = v_f$

hence, $$2mv = m_{wall} V$$

Also, $$p=mv$$

$$\frac{p}{m}=v$$

$$K= \frac{m}{2}v^2 = m \frac{p^2}{2m^2} = \frac{p^2}{2m}$$

Now if you do conservation of energy, you would notice that the wall infact gains some a little energy hence the collision is not perfectly elastic [ due to the fact that k.e is dependent on momentum in a way]

summary: The collision is only 'approximately' elastic

Answer 2

The change in momentum is due to mutual reaction force exerted by the wall and the ball on each other. Why? Because of Newton's second law, which states that,

$$\mathbf F = \frac{\mathrm d \mathbf p }{\mathrm d t}$$

which says that the rate of change of momentum is equal to the force applied.

And by Newton's third law, we know that the force exerted by the ball on the wall and the force exerted by the wall on the ball were equal and opposite, thus the momentum change in each case must be equal and opposite and thus the total momentum is conserved. So on one side, as you said, the momentum of the ball changes by 50, but at the same time, the momentum of the wall also changes by 50 however in the opposite direction.

And you are right in saying that both the wall and ball undergo acceleration because at the time of the collision, both were acted upon by their mutual and equal (and opposite) reaction force.

Answer 3

I know momentums definition (p=mv) and how to calculate problems involving it, but there are bits I don't understand.

Take a ball thrown at a wall for instance with mass value (5) and velocity (5). This would mean the total momentum of ball is (25). The wall is attached to the earth so the mass is huge and the wall doesn't move (momentum of (0)). Total momentum in system is now (25). This ball hits a wall and bounces back in the opposite direction.

The ball now has a momentum of (-25) so its moving in the opposite direction from which it started.

So far I agree.

The wall doesn't 'move' due to its high mass (and undetectable gain in velocity). THe total momentum of the system still has a magnitude of (25) but there has been a change in momentum of (50) for the ball.

Here I disagree.

Due to conservation of momentum, the total momentum after the collision is still ($25$) like it was before the collision. Now ($-25$) is from the ball and ($50$) is from the wall with the earth attached to it, giving a total of ($25$). However, because of the big mass of the earth ($6\cdot 10^{24}$ kg) the recoil velocity of the earth is very small, $$v_\text{earth}=\frac{p_\text{earth}}{m_\text{earth}}= \frac{50\text{ kgm/s}}{6\cdot 10^{24}\text{kg}}=8.3\cdot 10^{-24}\text{ m/s},$$ but it is not zero.