Why does the man on a horse get thrown forward when the horse stops?

  • Thread starter nilic1
  • Start date
  • Tags
    Strange
In summary, a man of mass 50kg is riding a horse of mass 450kg at a steady speed of 10m/s. When the horse stops suddenly, the man is thrown forwards with a resulting speed of 100m/s, according to the law of conservation of momentum. However, when considering the law of conservation of energy, it is surprising to find that the kinetic energy after is greater than that before. Various scenarios, such as the horse stopping by pressing against the ground or the man being attached to a compressed spring, are discussed to explain this phenomenon. It is concluded that while total kinetic energy can increase in a super-elastic collision, this is not the case when a horse stops. The interaction between the horse and
  • #1
nilic1
39
0

Homework Statement



A man of mass 50kg is riding a horse of mass 450kg at a steady speed of 10m/s. The horse stops suddenly and the man is thrown forwards. Find the speed with which the man is thrown forwards.

Applying law of conservation of momentum, the resulting speed is 100m/s. Poor man.
However, I checked whether K.E. is conserved or not in such a situation. To my surprise, the K.E. after is greater than that before!

Homework Equations



K.E = 1/2 m v^2

The Attempt at a Solution



According to the law of conservation of energy... energy can neither be created nor destroyed, so what is wrong here?
 
Physics news on Phys.org
  • #2
How does the horse stop? If there is an external force on the horse how can you conserve the momentum of system ?
 
  • #3
Not counting the interaction between the horse and the ground, you can model the horse-man system as a cart with a spring on it and a ball attached to the end of the spring. You keep the spring compressed with some thread and let the cart go with a certain speed. Suddenly you burn off the thread, so the spring is released. The impulse of the spring is just enough to stop the cart, and the ball will take away all the momentum. The KE of the ball will be greater than the original KE of the whole system, as the elastic energy of the spring is added.
I do not know what exactly the horse does with the man and with the ground but some kind of elastic energy can transform into the KE of the man.

ehild
 
  • #4
aim1732 said:
How does the horse stop? If there is an external force on the horse how can you conserve the momentum of system ?

Thanks for answering. This was an exam question and there is no further information. There is no other way of finding the resulting speed anyway. The horse just decides to stop abruptly lol. Does the K.E. calculation show that this an impossible situation in practice?
 
  • #5
ehild said:
Not counting the interaction between the horse and the ground, you can model the horse-man system as a cart with a spring on it and a ball attached to the end of the spring. You keep the spring compressed with some thread and let the cart go with a certain speed. Suddenly you burn off the thread, so the spring is released. The impulse of the spring is just enough to stop the cart, and the ball will take away all the momentum. The KE of the ball will be greater than the original KE of the whole system, as the elastic energy of the spring is added.
I do not know what exactly the horse does with the man and with the ground but some kind of elastic energy can transform into the KE of the man.

ehild

Thanks for your reply. So speaking about conservation of K.E does not make sense then? Does it make sense to speak about conservation of momentum because there are external forces involved here?
 
  • #6
If the horse stops by pressing against the ground(and there is no slipping) then energy and momentum will be conserved because although there is an external friction force it does not perform any work. This however is assuming the horse's back is smooth.
 
  • #7
Well, the horse would stop even without the man and then all the momentum and KE would be lost. During the stopping process, its hovers press against the ground and deform it, so there is some displacement and some time during this interaction: neither the momentum nor the energy is conserved when the interaction with Earth is not taken into account. The momentum is conserved if that scenario holds which I outlined in my previous post: that the man is attached with a compressed spring to the back of the horse, and this spring is suddenly released. :)
 
  • #8
Thanks ehild very interesting observation. After some more research I learned that total K.E. can actually increase and is called a super-elastic collision, where energy as you suggested might come from some stored potential or elastic energy.
 
Last edited:
  • #9
Well, it is true, but this is certainly not the case when a horse stops. I think, with a slippery back, with no horizontal interaction force between horse and man, the man just would keep its initial 10 m/s velocity while the horse stops.

ehild
 

1. What is a "strange K.E. problem"?

A "strange K.E. problem" refers to a situation in which the kinetic energy (K.E.) of an object does not follow the expected pattern or behavior. This could be due to external factors such as friction or air resistance, or it could indicate an error in calculations or measurements.

2. How is K.E. calculated?

K.E. is calculated using the formula K.E. = 1/2 * m * v^2, where m represents the mass of the object and v represents its velocity. This formula can be applied to both linear and rotational motion.

3. Can K.E. be negative?

Yes, K.E. can be negative. This typically occurs when there is a decrease in velocity, such as during deceleration or when work is done on an object to reduce its kinetic energy.

4. What is the relationship between K.E. and potential energy?

K.E. and potential energy are related through the principle of conservation of energy. As one form of energy decreases, the other form increases, and vice versa. For example, when an object is dropped, its potential energy decreases as it falls, while its kinetic energy increases.

5. How can a "strange K.E. problem" be solved?

A "strange K.E. problem" can be solved by carefully examining the situation and identifying any external factors that may be affecting the object's kinetic energy. It may also be helpful to double check calculations and measurements to ensure accuracy. In some cases, seeking the advice of a colleague or conducting further research may also lead to a solution.

Similar threads

  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
10K
  • Advanced Physics Homework Help
Replies
5
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
6K
  • Introductory Physics Homework Help
Replies
2
Views
5K
Replies
11
Views
5K
  • Introductory Physics Homework Help
Replies
7
Views
17K
Back
Top