Acceleration of two Objects after Collision

In summary: But Physics must relate to the real physical phenomena. It's not a random invention. Either a bike slides to a halt like a box or it rolls. And if it rolls, then kinetic friction is not acting. Otherwise, the wheel would have been a poor invention.My problem is not with the way you tackled the problem, but the people who set these questions divorce physics from reality and engineering.
  • #1
Riman643
62
2
Homework Statement
Two bikes collide on a street and travel a distance away from each other before coming to rest. The first bike is 20 kg and the second bike is 15 kg. The coefficient of kinetic friction is 0.5. What is the magnitude of acceleration of each.
Relevant Equations
F = ma
f[SUB]k[/SUB] = Nμ[SUB]k[/SUB]
This seems like a fairly simple problem but I got the same answer for both so I'm unsure if I am doing it right.

I found the accelerations as such

a1 = (fk)/(ma) = (98/20) = 4.9 m/s2
a2 = (fk)/(ma) = (73.5/15) = 4.9 m/s2

Am I doing something wrong or is this problem just that simple?
 
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  • #2
Riman643 said:
Problem Statement: Two bikes collide on a street and travel a distance away from each other before coming to rest. The first bike is 20 kg and the second bike is 15 kg. The coefficient of kinetic friction is 0.5. What is the magnitude of acceleration of each.
Relevant Equations: F = ma
fk = Nμk

This seems like a fairly simple problem but I got the same answer for both so I'm unsure if I am doing it right.

I found the accelerations as such

a1 = (fk)/(ma) = (98/20) = 4.9 m/s2
a2 = (fk)/(ma) = (73.5/15) = 4.9 m/s2

Am I doing something wrong or is this problem just that simple?
It could work out like that.

See if that makes sense.

What all goes into determining fk ?

What goes into getting N, the normal force, for this problem?
 
  • #3
I got the Normal force by knowing that it is equal to the force of gravity of each object. I then multiplied the normal force by the kinetic coefficient to get the force of friction.
 
  • #4
Riman643 said:
I got the Normal force by knowing that it is equal to the force of gravity of each object. I then multiplied the normal force by the kinetic coefficient to get the force of friction.
OK.
So, for motorcycle 1: (fk)1= μkN1 = μk(m1 g)
Then what do you do to find a1 ?
Divide that by m1. Right?
 
  • #5
SammyS said:
OK.
So, for motorcycle 1: (fk)1= μkN1 = μk(m1 g)
Then what do you do to find a1 ?
Divide that by m1. Right?
Correct. That's the only way I could think of to find the acceleration.
 
  • #6
Riman643 said:
Correct. That's the only way I could think of to find the acceleration.
It's a pretty good way.

Doing it symbolically:
## a_1=\dfrac{\mu_k (m_1 g)}{m_1}##

Does that show you that the acceleration of bike 1 is independent of its mass ?
 
  • #7
Riman643 said:
Correct. That's the only way I could think of to find the acceleration.
Do you think this problem makes sense? What has kinetic friction to do with a bicycle?
 
  • #8
SammyS said:
It's a pretty good way.

Doing it symbolically:
## a_1=\dfrac{\mu_k (m_1 g)}{m_1}##

Does that show you that the acceleration of bike 1 is independent of its mass ?
I think so. The mass on top would cancel with the mass on the bottom.
 
  • #9
Riman643 said:
I think so. The mass on top would cancel with the mass on the bottom.
Right.
So now, is it surprising that both bikes have the same magnitude of acceleration?
 
  • #10
PeroK said:
Do you think this problem makes sense? What has kinetic friction to do with a bicycle?
Kinetic friction would be the force slowing them down to a rest.
 
  • #11
SammyS said:
Right.
So now, is it surprising that both bikes have the same magnitude of acceleration?
Yeah now that I see it algebraically it makes more sense.
 
  • #12
Riman643 said:
Kinetic friction would be the force slowing them down to a rest.

1) Wouldn't a bike slow down somewhat and then fall over?

2) Why would kinetic friction slow a bike? If kinetic friction of ##0.5## was acting on a bike, then it would be almost impossible to get it to move.

3) E.g. you have calculated a deceleration of ##4.9m/s^2##. Does that seem realistic? A bicycle traveling at about ##5m/s## would come to a halt in ##1s## - almost as soon as you stop pedalling?
 
  • #13
PeroK said:
1) Wouldn't a bike slow down somewhat and then fall over?

2) Why would kinetic friction slow a bike? If kinetic friction of ##0.5## was acting on a bike, then it would be almost impossible to get it to move.

3) E.g. you have calculated a deceleration of ##4.9m/s^2##. Does that seem realistic? A bicycle traveling at about ##5m/s## would come to a halt in ##1s## - almost as soon as you stop pedalling?

As for the logistics of the question I am not sure. I think these were just randomly generated numbers. I am just curious if the way I went about solving the problem was correct.
 
  • #14
Riman643 said:
As for the logistics of the question I am not sure. I think these were just randomly generated numbers. I am just curious if the way I went about solving the problem was correct.

But Physics must relate to the real physical phenomena. It's not a random invention. Either a bike slides to a halt like a box or it rolls. And if it rolls, then kinetic friction is not acting. Otherwise, the wheel would have been a poor invention.

My problem is not with the way you tackled the problem, but the people who set these questions divorce physics from reality and engineering.

For your information, if you slam on the brakes on a bike (or a car) and it skids, then that is kinetic friction. If you brake properly, then that is actually static friction (although that takes a bit of explaining). And, if a bike or a car rolls to a stop without braking, then that is a combination of air resistance and rolling resistance. The fact that rolling resistance is so low compared to friction is why the wheel, bikes and cars are so efficient.

If it was all just friction then you might as well have square wheels on your car.
 

Related to Acceleration of two Objects after Collision

1. What is the formula for calculating the acceleration of two objects after a collision?

The formula for calculating the acceleration of two objects after a collision is a = (m1v1 + m2v2) / (m1 + m2), where m represents the mass of each object and v represents the initial velocity of each object.

2. Does the type of collision affect the acceleration of the objects?

Yes, the type of collision can affect the acceleration of the objects. In an elastic collision, both kinetic energy and momentum are conserved, resulting in a smaller change in acceleration compared to an inelastic collision where kinetic energy is not conserved.

3. How does the direction of the objects' velocities affect their acceleration after a collision?

The direction of the objects' velocities can affect their acceleration after a collision. In a head-on collision where the objects are moving towards each other, their accelerations will be in opposite directions. In a rear-end collision where one object is stationary and the other is moving towards it, their accelerations will be in the same direction.

4. Can the mass of the objects affect their accelerations after a collision?

Yes, the mass of the objects can affect their accelerations after a collision. The larger the mass of an object, the smaller its acceleration will be compared to an object with a smaller mass, assuming the same initial velocities and type of collision.

5. Is the acceleration of the objects after a collision the same for all collisions?

No, the acceleration of the objects after a collision can vary depending on the initial velocities, masses, and type of collision. In general, the acceleration will be smaller in elastic collisions compared to inelastic collisions, and will also depend on the directions of the objects' velocities.

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