Question

A 50 kg boy jumps off the front of a 1.5 kg skateboard moving forward. Find the skateboard’s velocity immediately after the boy jumps, assuming that the skateboard’s initial velocity is 3.5 m/s and the boy’s velocity when jumping off the front is 5 m/s.

Answer 1

the skate board velocity is -1.35 m/s

Step-by-step explanation:

Here we are goin to use the principle of conservation of linear momentum that states:

`m1*vo1+m2*vo2=m1*vf+1m2*vf2`

that is:

`(50kg+1.5kg)*3.5m/s=50kg*Vf1+1.5Kg*Vf2`

the velocity of the boy is 5m/s relative to the skate:

`Vbs=Vb-Vs\\Vb=Vbs+Vs=5m/s+Vf2`

`(50+1.5)*3.5=50*(5 +vf2)+1.5*Vf2\\180.25=250+(51.5)Vf2\\vf2=(-69.75)/(51.5)=-1.35m/s`

Answer 2

Here in this case we can say that there is no external force on skate board and boy as a system

so here we will use the theory of momentum conservation

So we will say

initial momentum of boy + board = final momentum of boy + board

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

`(50 + 1.5)(3.5) = 50(5 + v_2) + 1.5v_2`

`180.25 = 250 + 51.5 v_2`

`180.25 - 250 = 51.5 v_2`

`-69.75 = 51.5 v_2`

`v_2 = -1.35 m/s`

so here the skateboard will move off in opposite direction with speed 1.35 m/s