Question

A 45 kg boy runs at a rate of 2.5 m/s and jumps on top of a stationary skateboard that has a mass of 4 kg. After jumping onto the board, the boy and the board travel off together. Assuming that momentum is conserved, what is the final speed of the boy and the board?

Answer 1

V = 2.29 m/s

Step-by-step explanation:

Given that,

Mass of the boy,
`m_1=45\ kg`

Mass of the skateboard,
`m_2=4\ kg`

Initial speed of the boy, v = 2.5 m/s

Let V is the final velocity of the boy and the board. The net momentum of the system remains constant. Using the conservation of linear momentum to find it as :

`45* 2.5=(45+4)V`

`V=(45* 2.5)/((45+4))`

V = 2.29 m/s

So, the velocity of the boat after Batman lands in it 2.29 m/s. Hence, this is the required solution.

Answer 2

v = 2.29 m/s

Step-by-step explanation:

As we know that the external force on the system of mass of boy + board is ZERO

So here we can use momentum conservation

now we have

`m_1v_1 + m_2v_2 = (m_1 + m_2) v`

now we have

`45 (2.5) + 4(0) = (45 + 4) v`

now we have

`v = (45)/(49) (2.5)`

`v = 2.29 m/s`