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Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his cart, mcart, is identical to the combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest. Chuck then picks up a ball of mass mball and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground (ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is vc. The speed of the thrown ball relative to the ground is vb. Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie's speed relative to the ground after she catches the ball is vj. When answering the questions in this problem, keep the following in mind: The original mass mcart of Chuck and his cart does not include the mass of the ball. The speed of an object is the magnitude of its velocity. An object's speed will always be a nonnegative quantity.

Respuesta :

The Question is not complete and this is the complete question;

Part A

Find the relative speed u between Chuck and the ball after Chuck has thrown the ball.

Express the speed in terms ofv_c and v_b.

Part B

What is the speed v_b of the ball (relative to the ground) while it is in theair?

Express your answer in terms of m_ball, Image for Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck, and u.

Part C

What is Chuck's speed v_c (relative to the ground) after he throws theball?

Express your answer in terms ofm_ball, Image for Chuck and Jackie stand on separate carts, both of which canslide without friction. The combined mass of Chuck, and u.

Part D

Find Jackie's speed v_j (relative to the ground) after she catches the ball, in terms of v_b.

Express v_j in terms of m_ball, Image for Chuck and Jackie stand on separate carts, both of which canslide without friction. The combined mass of Chuck, and v_b.

Part E

Find Jackie's speed v_j (relative to the ground) after she catches theball, in terms of u.

Express v_j in terms of m_ball, Image for Chuck and Jackie stand on separate carts, both of which canslide without friction. The combined mass of Chuck, and u.

Answer:

A) u = Vc + Vb

B) Vb = m(cart)u/ [(mball) + m(cart)]

C) Vc = um(ball)/[(mball) + m(cart)]

D) Vj = m(ball)Vb/ [(mball) + m(cart)]

E) Vj = (m(ball) x um(cart)) / [(mball) + m(cart)]²

Explanation:

A) We want to find the relative speed between chuck and the ball after he has thrown the ball.

Now, chuck and the ball are opposite to each other , therefore the relative speed between them is equal to the sum of their speeds, so; u = chucks speed + ball's speed

u = Vc + Vb

B) From conservation of momentum, we know that;

Initial momentum = Final momentum.

Since balls are at rest, initial momentum(P1) is zero

Now, final momentum (P2) is;

P2 = - m(cart)Vc + m(ball)Vb

Now since P1 = P2 and P1 =0.thus,

- m(cart)Vc + m(ball)Vb = 0

Add - m(cart)Vc to both sides to obtain;

m(ball)Vb = m(cart)Vc

Vb = (m(cart)Vc) / m(ball)

From answer a above, Vc = u - Vb

So, Vb = [m(cart)(u - Vb] / m(ball)

Multiply both sides by m(ball) to get;

Vb(mball) = m(cart)u - m(cart)Vb

Add m(cart)Vb to both sides to get;

Vb(mball) + m(cart)Vb = m(cart)u

Vb[(mball) + m(cart)] = m(cart)u

So, Vb = m(cart)u/ [(mball) + m(cart)]

C) From above; Vc = u - Vb

So, Vc = u - m(cart)u/ [(mball) + m(cart)]

Vc = [u((mball) + m(cart))] - m(cart)u] / [(mball) + m(cart)]

So, Vc = um(ball)/[(mball) + m(cart)]

D) In this case initial momentum = m(ball)Vb while final momentum is [(mball) + m(cart)]Vj

Thus as stated in b above, in conservation of momentum,

Initial momentum = final momentum.

Hence ;

m(ball)Vb = [(mball) + m(cart)]Vj

So; Vj = m(ball)Vb/ [(mball) + m(cart)]

E) From earlier, we saw that;

Vb = m(cart)u/ [(mball) + m(cart)]

So putting that for Vb in D above to get;

Vj = [m(ball)/ [(mball) + m(cart)]] x [m(cart)u/ [(mball) + m(cart)]

= (m(ball) x um(cart)) / [(mball) + m(cart)]²

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