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

Answer:

v = 2.29 m/s

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

$v = \frac{45}{49} (2.5)$

$v = 2.29 m/s$

Answer 2

Answer:

V = 2.29 m/s

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 :

$45\times 2.5=(45+4)V$

$V=\dfrac{45\times 2.5}{(45+4)}$

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.