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LiRa [457]
3 years ago
9

A 50 kg woman, riding on a 10 kg cart, is moving east at 5.0 m/s. the woman jumps off the cart and hits the ground running at 7.

0 m/s, eastward, relative to the ground. calculate the velocity of the cart after she jumps off.
Physics
1 answer:
pav-90 [236]3 years ago
4 0
To answer this question, we will use the law of conservation of momentum which states that:
(m1+m2)Vi = m1V1 + m2V2 where:
m1 is the mass of the woman = 50 kg
m2 is the mass of the cart = 10 kg
Vi is the initial velocity (of woman and cart combined) = 5 m/sec
V1 is the final velocity of the woman = 7 m/sec
V2 is the final velocity of the cart that we need to calculate

Substitute with the givens in the above equation to get the final velocity of the cart as follows:
(50+10)(5) = (50)(7) + (10)V2
10V2 = -50
V2 = -5 m/sec
Note that the negative sign indicates that the cart is moving in an opposite direction to the others.
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Answer:

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Explanation:

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substituting values

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      V_w = \frac{V}{2}  

substituting values

     V_w = \frac{1162.7 }{2}

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     W_r  =  49.0 \ N

The tension on the string is

       T  = W_r - W_w

substituting values

       T  = 49.0 -  5.698

       T  =  43.302 \ N

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