Answer:
[tex]v=3.564\ m.s^{-1}[/tex]
[tex]\Delta v =2.16\ m.s^{-1}[/tex]
Explanation:
Given:
(A)
height between John and William, [tex]h=1.8\ m[/tex]
Using the equation of motion:
[tex]v_J^2=u_J^2+2 (g.sin\theta).l[/tex]
where:
v_J = final velocity of John at the end of the slide
u_J = initial velocity of John at the top of the slide = 0
Now putting respective :
[tex]v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3[/tex]
[tex]v_J=5.94\ m.s^{-1}[/tex]
Now using the law of conservation of momentum at the bottom of the slide:
Sum of initial momentum of kids before & after collision must be equal.
[tex]m_J.v_J+m_w.v_w=(m_J+m_w).v[/tex]
where: v = velocity with which they move together after collision
[tex]30\times 5.94+0=(30+20)v[/tex]
[tex]v=3.564\ m.s^{-1}[/tex] is the velocity with which they leave the slide.
(B)
Now we find the force along the slide due to the body weight:
[tex]F=m_J.g.sin\theta[/tex]
[tex]F=30\times 9.8\times \frac{1.8}{3}[/tex]
[tex]F=176.4\ N[/tex]
Hence the net force along the slide:
[tex]F_R=71.4\ N[/tex]
Now the acceleration of John:
[tex]a_j=\frac{F_R}{m_J}[/tex]
[tex]a_j=\frac{71.4}{30}[/tex]
[tex]a_j=2.38\ m.s^{-2}[/tex]
Now the new velocity:
[tex]v_J_n^2=u_J^2+2.(a_j).l[/tex]
[tex]v_J_n^2=0^2+2\times 2.38\times 3[/tex]
[tex]v_J_n=3.78\ m.s^{-1}[/tex]
Hence the new velocity is slower by
[tex]\Delta v =(v_J-v_J_n)[/tex]
[tex]\Delta v =5.94-3.78= 2.16\ m.s^{-1}[/tex]
