Chuck and Jackie stand on separate carts, both of which can slide without fricti

Question

Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his cart, , 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 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 . The speed of the thrown ball relative to the ground is .
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 .
When answering the questions in this problem, keep the following in mind:
The original mass 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.

Explanation / Answer

a) Due to momentum conservation chuck and ball will move in opposite direction.

Thus, relative velocity u = v_c + v_b

b) Momentum conservation: m_cart*0 = m_cart*v_c - m_b*v_b

m_b*v_b = m_cart*v_c

v_b = (m_cart/m_b)*(u - v_b)

u/v_b - 1 = m_b/m_cart

v_b = u*m_cart/(m_b + m_cart)

c) v_c = (m_b/m_cart)*v_b

v_c = (m_b/m_cart) * u*m_cart/(m_b + m_cart)

v_c = u*m_b/(m_b + m_cart)

d) Momentum conservation: m_cart*0 + m_b*v_b = (m_cart + m_b)*v_j

v_j = v_b*m_b/(m_cart + m_b)

e) v_j = u*m_cart/(m_b + m_cart) *m_b/(m_cart + m_b)

v+j = u*m_b*m_cart/(m_cart + m_b)^2

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