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gogolik [260]
3 years ago
5

A 35.4 kg girl is riding on the back of a 15.23 kg cart. the cart and the kid are both moving eastward at 4.25 m/s when she step

s off the back of the cart. if the final velocity of the girl is 3.06 m/s east ward, what is the final speed of the cart?
Physics
1 answer:
Grace [21]3 years ago
7 0

Answer:

The final  velocity of the cart is  v_c = 7.02 \  m/s

Explanation:

From the question we are told that

    The mass of the girl is  m_g  = 35.4 \ kg

     The mass of the cart is  m_c  = 15.23 \ kg

      The speed of the cart and  kid(girl) is  v = 4.25 \ m/s

     The final velocity of  the girl is v_g  = 3.06 \  m/s

Let assume that velocity eastward is  positive and velocity westward is negative (Note that if we assume vise versa it wouldn't affect the answer )

   The total momentum of the system before she steps off the back of the cart

is mathematically evaluated as

        p__{T1}} = (m_g + m_c) * v

substituting values

        p__{T1}} = (35.4 + 15.23) * 4.25

        p__{T1}} =215.17 \  kg m /s

The total momentum after she steps off the back of the cart is mathematically evaluated as

        p__{T2}} = (m_g * v_g ) +(  m_c * v_c )

Where  v_c  is the final velocity of the cart

substituting values    

      p__{T2}} = (35.4 * 3.06 ) +(  15.23 * v_c )

       p__{T2}} = 108. 324 + 15.23  v_c

Now according to the law of conservation of momentum

       p__{T1}} =p__{T2}}

So  

       215.17 \  kg m /s =  108. 324 + 15.23  v_c

=>      v_c = 7.02 \  m/s

Since the value is positive it implies that the cart moved eastward

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