Ambiguous question on momentum and how it works...

[jeff einstein]

Homework Statement

A question states that a ball with a mass of 5 kg and a velocity of 10 m/s collides with a stationary wall. The question then asks for the momentum of the ball after the collision.

Relevant Equations

P=mv, f=(dleta)p/t

A question states that a ball with a mass of 5 kg and a velocity of 10 m/s collides with a stationary wall. The question then asks for the momentum of the ball after the collision. No other information is given. This is a rather ambiguous question, as it doesn't provide the time taken for the collision or other necessary variables. Initially, the wall is stationary and therefore has no velocity or momentum. According to the conservation of momentum, the total momentum of the system must remain constant after the collision.

I know for certain that the ball will move in the negative direction after the collision (assuming the initial direction of motion is positive). Given this, I initially thought that the wall's velocity would remain zero. However, if this were true, the ball wouldn’t move in the negative direction. After some deep thought and confusion, I checked the mark scheme for this question. It stated that the ball would have a momentum of the same magnitude (50 kg·m/s) but in the opposite direction. The reasoning provided wasn’t very clear. It mentioned that momentum is a vector, so it changes from positive to negative, and the magnitude remains the same because the velocity and mass don’t change.

This answer still didn’t provide me with clarity, and I feel that the mark scheme is wrong. The magnitude of the ball's momentum can be the same, but this is just one of many infinite possibilities. The ball's change in momentum depends on the time taken for the collision, and the wall doesn’t remain stationary after the collision. It gains some momentum, but because it is so massive (since it's connected to the Earth), the velocity change is extremely small. This allows us to safely call the wall stationary, even though it technically has some velocity due to the momentum transferred from the ball.

My question now is: Can the ball’s momentum in the negative direction have a greater magnitude, or does it remain unchanged (as the mark scheme suggests) and what factors does this depend on?

 

[Orodruin]

Conservation of momentum only holds if there are no external forces acting on the system. In this case there is an external force from the stationary wall on the ball. The only piece of information missing is whether the collision is elastic or not, and if not what the coefficient of restitution is. Based on the expected answer, the collision is assumed elastic, which means energy is conserved.

If you promptly want to model the motion of the wall, you can give it a large mass $M$ (or consider the wall attached to the Earth, in which case that mass is the mass of the Earth), which will lead to very negligible motion of the wall.

 

[haruspex]

. The only piece of information missing is whether the collision is elastic

.. and the angle at which the ball strikes the wall?

If you promptly want to model the motion

"properly"?

 

[jeff einstein]

Lets say the collision is perfectly elastic and the angle at which the ball strikes is perpendicular to the flat wall

 

[BvU]

Why make things difficult (if you can make them impossible with just a little effort ). ?

Turn it around: if the exercise composer actually intended something more complicated (than 100 % coefficient of restitution) he/she would have added more information.

$\$

 

[kuruman]

This is a rather ambiguous question, as it doesn't provide the time taken for the collision or other necessary variables.

I agree that this question is ambiguous but I disagree with the reason you provide for the ambiguity. The momentum change of the ball is related to the average force exerted during the collision and the time taken for the collision by the $$m(v_{\!f}-v_{i})=F_{\text{avg.}}\Delta t.$$ Here, you are given $m=5~$kg, $v_i=10~$m/s and you are asked to find the unknown $v_{\!f}$. Solving the impulse equation for it, $$v_{\!f}=v_i+\frac{F_{\text{avg.}}}{m} \Delta t.$$ Even if the problem provided a numerical value for the collision time $\Delta t$, you still would not be able to find a numerical value for the final velocity because you lack a value for the average force.

To answer this problem with a numerical answer, you need more numerical information. This could be the average force and the collision time. The need for these two can be bypassed if a numerical value for the coefficient of restitution is provided as noted by others.

 

[kuruman]

Why make things difficult (if you can make them impossible with just a little effort ). ?

Turn it around: if the exercise composer actually intended something more complicated (than 100 % coefficient of restitution) he/she would have added more information.

$\$

I believe that this is dangerous thinking because it can backfire. It encourages people to invent their own information if they believe that not enough is given. What if they actually have what they need but don't realize it?

 

[gmax137]

Homework Statement: A question states that a ball with a mass of 5 kg and a velocity of 10 m/s collides with a stationary wall. The question then asks for the momentum of the ball after the collision.
Relevant Equations: P=mv, f=(dleta)p/t

My question now is: Can the ball’s momentum in the negative direction have a greater magnitude

Do you mean, could the ball's velocity be more than 10 m/sec (in the negative direction)? I don't think so (What would that imply?)

EDIT:
Maybe this post provides clues
https://www.physicsforums.com/threa...-e-1-2mv-2-more-e-with-v.1067128/post-7136115

 

[kuruman]

I have seen problems in which a firecracker or a cocked spring release energy at the point of the collision so that the final kinetic energy of the colliding masses is greater than the initial.

 

[gmax137]

I have seen problems in which a firecracker or a cocked spring release energy at the point of the collision so that the final kinetic energy of the colliding masses is greater than the initial.

That's why I said

I don't think so (What would that imply?)

Instead of just "No."

 

[kuruman]

That's why I said

Instead of just "No."

You did, but "I don't think so" sometimes implies "No" as in

- I am going to the movies and you are coming with me.​

- I don't think so.​

 

[Lnewqban]

The magnitude of the ball's momentum can be the same, but this is just one of many infinite possibilities.

The less elastic the collision, the less the rebound velocity and the more the amount of kinetic energy that goes into heating the ball and wall.

This allows us to safely call the wall stationary, even though it technically has some velocity due to the momentum transferred from the ball.

Consider that the Earth-wall velocity was initially affected by the throw of the ball, just to go back to the initial condition after the impact.

My question now is: Can the ball’s momentum in the negative direction have a greater magnitude...?

Only if additional energy is added to the ball during the collision.
Example: the velocity of a baseball is greater after than before colliding with a bat that has muscles swinging it.

 

[Orodruin]

Example: the velocity of a baseball is greater after than before colliding with a bat that has muscles swinging it.

This is due to the velocity of the bat and hence the colliding object not being at rest. In fact, the collision happens so fast that the actual energy put into the ball by the muscles during collision is effectively zero. As a batter you may not feel the collision until after it is over as that disturbance has to propagate through the bat before it reaches your hands.

 

[Mister T]

Homework Statement: A question states that a ball with a mass of 5 kg and a velocity of 10 m/s collides with a stationary wall. The question then asks for the momentum of the ball after the collision.
Relevant Equations: P=mv, f=(dleta)p/t

I know for certain that the ball will move in the negative direction after the collision (assuming the initial direction of motion is positive).

It's possible that the ball could stop after the collision (perfectly inelastic).