Is momentum conserved in y direction ?

[bksree]

Hi
A person standing on a cart throws a ball at an angle to the horizontal
Is momentum conserved
(i) in x direction if the cart is on a frictionless surface ?
(ii) in x direction if cart is on real ground (i.e. with friction)
(iii) in y direction

My answer is : with a system consisting of man, cart and ball
(i) yes (ii) no because ground exerts frictional force on cart
(iii) yes because weight of man + cart + any reaction exerted by him is balance by ground and hence external force in y direction is zero again

TIA

 

[mfb]

(iii) yes because weight of man + cart + any reaction exerted by him is balance by ground and hence external force in y direction is zero again

"Balanced by ground" means that there is an external force. Just imagine an angle of pi/2, so the man is throwing the ball vertically. Does the ball get a vertical momentum component? Does the man or the cart get one? Therefore, is momentum conserved in this system?

 

[bksree]

Thanks. My doubt is :
When the cart, man & ball are stationary, the external forces acting on the system of cart, man & ball is the weight Mg vertically downwards and the normal reaction vertically upwards. The net vertical external force is zero.

When the man throws the ball at an angle θ to the ground, the vertical forces on the system of man + cart is :
Mg (downwards)
reaction of ball (downwards)
normal reaction of ground (upwards)

This is again balanced and net external force is zero. Hence shouldn't momentum be conserved in y direction

TIA

 

[mfb]

When the man throws the ball at an angle θ to the ground, the vertical forces on the system of man + cart is :
Mg (downwards)
reaction of ball (downwards)
normal reaction of ground (upwards)

This is again balanced and net external force is zero. Hence shouldn't momentum be conserved in y direction

The forces on man+cart are balanced, but the force on the ball is not. Therefore, the force on man+ball+cart together is not balanced (as long as the man is accelerating the ball).

 

[siddharth23]

Thanks. My doubt is :
When the cart, man & ball are stationary, the external forces acting on the system of cart, man & ball is the weight Mg vertically downwards and the normal reaction vertically upwards. The net vertical external force is zero.

When the man throws the ball at an angle θ to the ground, the vertical forces on the system of man + cart is :
Mg (downwards)
reaction of ball (downwards)
normal reaction of ground (upwards)

This is again balanced and net external force is zero. Hence shouldn't momentum be conserved in y direction

TIA

The initial momentum in the y-direction was 0. After throwing the ball, the resultant momentum in the y-direction would be:
m_ball x component of velocity in the y direction

Is it conserved?

 

[bksree]

The forces on man+cart are balanced, but the force on the ball is not. Therefore, the force on man+ball+cart together is not balanced (as long as the man is accelerating the ball).

Thanks but i didn't get the explanation.
In my understanding

In X direction
man applies force on ball, ball accelerates, reaction of ball acts on man + cart. In system of man + ball + cart there is no external force (because friction is 0) and so momentum is conserved.

In Y direction
man applies force on ball, ball accelerates, reaction of ball acts on man + cart. In system of man + ball + cart the force in vertical direction is non zero but balanced (mg + ball reaction - N) hence shouldn't momentum be conserved ?

 

[BruceW]

Let's say there is just a ball, falling under gravity. Is momentum in the y direction conserved in this case?

 

[ehild]

The question "is momentum conserved" is wrong without specifying the object or system.

The momentum of a closed system is conserved. The change of momentum is equal to the impulse.

Is the momentum conserved then or does it change? It depends on the object or system of objects we speak about.

Throwing a ball up will conserve the momentum of the system ball-man-Earth, but the momentum of the ball changes. Internal forces do not change the momentum, but defining the system in different way, the internal force might become external.

So what is the question, the momentum of what is conserved?

ehild

 

[siddharth23]

Let's make this simple. For the ball man cart system (initially momentum 0 in the y direction), after the ball is thrown, it a y component in it's velocity. Do the man and cart have a y component?

 

[Millacol88]

Say that you take the Ball, Man, Cart system. The forces between the ball and the man, and the man and the cart are all action-reaction pairs and thus cancel out. But the cart also pushes on something outside of the system: the Earth. The Earth pushes back with a reaction force. Is this force not external to the system as you have chosen it? Then momentum is not conserved.

Now consider the Ball, Man, Cart, Earth system. Let's ignore the force of gravity from the Sun. Then the net external force on this system is 0. The man exerts a force on the ball, in turn the ball pushes on the man, and there are of course internal forces between the man, cart, and earth. The man, the cart and the Earth are imparted the same momentum that the ball has but in an opposite direction. So momentum is conserved.

So you see, the question of whether momentum is conserved is kind of meaningless unless you carefully specify the system.

 

[bksree]

Thanks to all for the replies. I agree that you have to specify the system clearly before asking the question. In this case the system is cart + man + ball together.

Say that you take the Ball, Man, Cart system. The forces between the ball and the man, and the man and the cart are all action-reaction pairs and thus cancel out. But the cart also pushes on something outside of the system: the Earth. The Earth pushes back with a reaction force. Is this force not external to the system as you have chosen it? Then momentum is not conserved.

But the NET external force on the system of ball+man+cart is zero i.e. N-mg = 0. Hence shouldn't momentum of this system be conserved

 

[Millacol88]

But the NET external force on the system of ball+man+cart is zero i.e. N-mg = 0. Hence shouldn't momentum of this system be conserved

The gravitational force and normal force from the Earth do not cancel out though. There is a reaction force from the Earth equal in magnitude to the force applied to the ball by the man. This force is not canceled out by the force of the cart pushing on the Earth because the Earth is not in your chosen system.

 

[bksree]

The gravitational force and normal force from the Earth do not cancel out though. There is a reaction force from the Earth equal in magnitude to the force applied to the ball by the man. This force is not canceled out by the force of the cart pushing on the Earth because the Earth is not in your chosen system.

?
Initially the external force acting on man+earth+ball = - mg and ground reaction = + N, and N-mg = 0.
When ball is thrown if force applied on ball is F, then total downward force on cart+man+ball = -mg - F and Earth reaction = N + ΔN where ΔN = F. If this weren't the case the cart would not remain on the ground, isn't it ?

 

[Millacol88]

?
Initially the external force acting on man+earth+ball = - mg and ground reaction = + N, and N-mg = 0.
When ball is thrown if force applied on ball is F, then total downward force on cart+man+ball = -mg - F and Earth reaction = N + ΔN where ΔN = F. If this weren't the case the cart would not remain on the ground, isn't it ?

Is it? If the ball is in your system then you must account for the upward force F applied on it.

 

[BruceW]

?
Initially the external force acting on man+earth+ball = - mg and ground reaction = + N, and N-mg = 0.
When ball is thrown if force applied on ball is F, then total downward force on cart+man+ball = -mg - F and Earth reaction = N + ΔN where ΔN = F. If this weren't the case the cart would not remain on the ground, isn't it ?

Think about what happens after the ball leaves the man's hand. F is not important. What is the gravitational force on the cart+man+ball system and what is the normal force acting on this system?

 

[ehild]

?
Initially the external force acting on man+earth+ball = - mg and ground reaction = + N, and N-mg = 0.
When ball is thrown if force applied on ball is F, then total downward force on cart+man+ball = -mg - F and Earth reaction = N + ΔN where ΔN = F. If this weren't the case the cart would not remain on the ground, isn't it ?

Let be the mass of ball m, and M is that of the cart with the man. Before the ball is thrown the normal force between the ground and cart is N=(m+M)g.
The man throws the ball with force F.
During the throw, the man and gravity exert forces on the ball. Ʃf(ball)= F-mg

The ball acts with force -F on the man-cart system. The ground acts with an increased normal force N+ΔN. So the net force on the man-cart is Ʃf(man-cart)= -F-Mg+N+ΔN=0 as it is unmoved.

The total force on the whole system ball-man-cart is F-mg-F-Mg+N+ΔN=ΔN.

F=(m+M)g-Mg+ΔN=mg+ΔN.
The ball experiences ΔN net force.

ehild

 

[bksree]

Thanks to all for the time and patience.

Let be the mass of ball m, and M is that of the cart with the man. Before the ball is thrown the normal force between the ground and cart is N=(m+M)g.
The man throws the ball with force F.
During the throw, the man and gravity exert forces on the ball. Ʃf(ball)= F-mg

The ball acts with force -F on the man-cart system. The ground acts with an increased normal force N+ΔN. So the net force on the man-cart is Ʃf(man-cart)= -F-Mg+N+ΔN=0 as it is unmoved.

The total force on the whole system ball-man-cart is F-mg-F-Mg+N+ΔN=ΔN.

F=(m+M)g-Mg+ΔN=mg+ΔN.
The ball experiences ΔN net force.

ehild

Thank you for the lucid explanation.

Is it? If the ball is in your system then you must account for the upward force F applied on it.

Now, I get your point. Thanks

 

[siddharth23]

We always consider an Earth frame of reference.

What's the final inference you drew?

 

[bksree]

Momentum is not conserved in Y direction