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Rus_ich [418]
3 years ago
11

Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical sur

face that is firmly attached to the cart. If the ball bounces straight back as shown in the picture, will the cart be put into motion after the ball bounces back from the surface?
A.Yes, and it moves to the right.
B.Yes, and it moves to the left.
C.No, it remains in place

Physics
1 answer:
Len [333]3 years ago
3 0

Answer:

F_c t_ c = -F_b t_b

And the forces are equal but in the opposite direction. So then we can write by general rule:

m_c \Delta V_{c} = -m_b \Delta V_b

Or equivalently:

m_c \Delta V_{c} +m_b \Delta V_b =0

Where: V_c represent the speed of the car and V_b the speed of the ball

m_c represent the mass of the car

m_b represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

Explanation:

If we assume that we have the situation in the figure attached.

For this case we assume that the momentum changes are equal in magnitude and opposite in direction, so then we satisfy this:

F_c t_ c = -F_b t_b

And the forces are equal but in the opposite direction. So then we can write by general rule:

m_c \Delta V_{c} = -m_b \Delta V_b

Or equivalently:

m_c \Delta V_{c} +m_b \Delta V_b =0

Where: V_c represent the speed of the car and V_b the speed of the ball

m_c represent the mass of the car

m_b represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

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Answer:

P1 = 1.3 (500 + 60) = 728 kg-m      total momentum to right at start

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total momentum after running at -10 with respect to cart = 728 where v2 is the new speed of the cart

728 = 560 v2 - 600

v2 = 1328 / 560 = 2.37 m/s    new speed of cart

Check:

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p1 for man = (2.37 - 10) * 60 = -458

P2 = p1 + p2 = 728       total momentum unchanged

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