A 50kg woman, riding a 10kg cart, is movingeast @ 5.0m/s. tha woman jumps off th

Question

   A 50kg woman, riding a 10kg cart, is movingeast @ 5.0m/s. tha woman jumps off the cart and hit the ground @7m/s eastward, relitive to the ground. Find the velocity of the cart after the woman jumps off. a) 50kg+10kg = m1    5.0m/s         = v1     300kgm/s    = p1 b)50kg            =m2       7.0m/s        =v2       350kg m/s = p2 Inthe opp direction 300 +(-350)= -50p v =-50/10 v = -5 m/s Velocity of the cart is 5 m/s in a westerly direction.    A 50kg woman, riding a 10kg cart, is movingeast @ 5.0m/s. tha woman jumps off the cart and hit the ground @7m/s eastward, relitive to the ground. Find the velocity of the cart after the woman jumps off. a) 50kg+10kg = m1    5.0m/s         = v1     300kgm/s    = p1 b)50kg            =m2       7.0m/s        =v2       350kg m/s = p2 Inthe opp direction 300 +(-350)= -50p v =-50/10 v = -5 m/s Velocity of the cart is 5 m/s in a westerly direction.

Explanation / Answer

Use the principle of conservation of momentum. (We'll have toassume that the motion was not affected by outside forces likefriction or gravity; because momentum is not conserved if there areoutside forces acting on the system.) For velocity, let's use positive numbers for "east" and negativenumbers for "west". (This is arbitrary; could easily choose theopposite without affecting the answer.) Total momentum before the jump: (mass of woman & cart) × (velocity of woman &cart) (50kg+10kg)(5.0m/s) Total momentum after the jump: (m_woman)(v_woman) + (m_cart)(v_cart) (50kg)(7.0m/s) + (10kg)(v_cart) Momentum is conserved which means: Total momentum before = total momentum after (50kg+10kg)(5.0m/s) = (50kg)(7.0m/s) + (10kg)(vcart) Now just use algebra to solve for "vcart": vcart = ((50kg+10kg)(5.0m/s) (50kg)(7.0m/s) ) /10kg = 5.0m/s. Since we said that negative velocities mean "west," this means thatthe direction of vcart is westward with velocity 5.0m/s.

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