Why momentum of a ball bounced off a wall increases twice fold?

[Eobardrush]

Homework Statement

This is for the end of chapter homework from a textbook(I do self study)

Relevant Equations

P=Mv

Delta P= mv-mu

Question 2a: It is really hard for me to get my head around this.

The solution of this question mentions the momentum of the ball after it rebounds is 12kgms. My attempt at this solution is as follows

Before collision

Momentum of ball= mv= 2x3= 6 kgms and momentum of wall= 0

Therefore Total momentum of system before collision= 6+0= 6kgms

But since momentum of the system is conserved that means the total momentum of the system before collision should equal to the total momentum of the system after collision

And since the total momentum before collision is 6 kgms, shouldn't the total momentum after collision should be equal to 6kgms too??
I don't understand how the balls momentum suddenly increases twice fold to 12kgms

I do understand that change in momentum equation mv-(-mv) results in 12kgms but it still is bugging me because I really do not get why the ball momentum suddenly increases to 12kgms despite it being at the same speed and mass. It kind of gives the sense that the momentum suddenly increased out of nowhere

 

[mfb]

Momentum has a direction, watch that and the signs of the velocity.

The total momentum after the collision is the same as before but the wall receives some momentum, that means the momentum of the ball can change (it's inevitable from the change in direction). The final momentum of the ball is -6 kg m/s and the change (which the question asks about) is 12 kg m/s.

 

[haruspex]

The total momentum after the collision is the same

... in magnitude ...

 

[Eobardrush]

The total momentum after the collision is the same as before but the wall receives some momentum, that means the momentum of the ball can change (it's inevitable from the change in direction). The final momentum of the ball is -6 kg m/s and the change (which the question asks about) is 12 kg m/s.

Thank you very much for the reply! Really helped a lot.
I do want to confirm one thing tho

So even if the ball hits the wall at 3 m/s and rebounds back with the same speed of 3 m/s, the ball still shared some of its momentum with the wall. But the momentum shared with the wall is so small that the change in the magnitude of the velocity is not even noticeable right? Which is why the ball still have the same speed of 3 m/s even when it did rebound off the wall

 

[Steve4Physics]

And since the total momentum before collision is 6 kgms, shouldn't the total momentum after collision should be equal to 6kgms too??
I don't understand how the balls momentum suddenly increases twice fold to 12kgms

Hi @Eobardrush and welcome to PF. I’ll chip-in too.

kgms should be kgm/s or kgms⁻¹

The question asks for the change in momentum, not the final momentum.

In this problem we take the initial direction as the positive direction (since we are given a positive value for the initial velocity). The ‘backwards’ direction is the negative direction. That means the final velocity is -3.0ms⁻¹. Note the minus sign.

change = (final value) – (initial value)

The official answer (12kgms⁻¹) is wrong. It should be -12kgms⁻¹. Note the minus sign. (Though the magnitude of the change is 12kgms⁻¹.)

The momentum of a system is only conserved if no external forces act on the system. If the system is the ball only, then there was an external force on it. It is interesting to ask what happens to (the wall+rest of the earth) when the ball bounces off the wall!

 

[PeroK]

Thank you very much for the reply! Really helped a lot.
I do want to confirm one thing tho

So even if the ball hits the wall at 3 m/s and rebounds back with the same speed of 3 m/s, the ball still shared some of its momentum with the wall. But the momentum shared with the wall is so small that the change in the magnitude of the velocity is not even noticeable right? Which is why the ball still have the same speed of 3 m/s even when it did rebound off the wall

The key point is that momentum is a vector quantity. It has direction. Momentum is $m\vec v$, where $\vec v$ is the velocity. Speed, on the other hand, is the magnitude of velocity, so $mv \equiv m|\vec v|$ is the magnitude of the momentum.

You must always be clear whether you are using speed or velocity and momentum or magniture of momentum.

If we consider the system here to be the ball, then momentum (of the ball) is not conserved. Nor, for example, is the momentum of a tennis ball during a rally.

 

[PeroK]

PS for one-dimensional motion, where you may want to use $v$ for the velocity, where $v$ may be positive or negative, you should write $|v|$ for the speed.

 

[jbriggs444]

So even if the ball hits the wall at 3 m/s and rebounds back with the same speed of 3 m/s, the ball still shared some of its momentum with the wall. But the momentum shared with the wall is so small that the change in the magnitude of the velocity is not even noticeable right?

The change in momentum of the ball was -12 kg m/s. [Change = final minus initial]

By Newton's third law or by conservation of momentum, the change in the momentum of the wall was equal and opposite: +12 kg m/s.

The reason that the wall did not move appreciably is that its mass is large and it is anchored to the Earth whose mass is larger still. If you divide +12 kg m/s by a bazillion metric tons, you get a negligible velocity.

 

[PeroK]

bazillion metric tons

It's $6$ zetta (sextillion) tons, apparently. Or $6$ yotta (septillion) kilograms.

 

[jbriggs444]

"Bazillion" is, of course, a technical term meaning "large enough that I don't want to count zeroes any more".