Question

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 steps 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?

Answer 1

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

Step-by-step 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