Respuesta :
Answer:
a) [tex]v=5.6725\,m.s^{-1}[/tex]
b) [tex]h= 1.6420\,m[/tex]
Explanation:
Given:
- mass of the body, [tex]M=20\,kg[/tex]
- mass of the tyre,[tex]m=10\,kg[/tex]
- length of hanging of tyre, [tex]l=3.5m[/tex]
- distance run by the body, [tex]d=10m[/tex]
- acceleration of the body, [tex]a=3.62m.s^{-2}[/tex]
(a)
Using the equation of motion :
[tex]v^2=u^2+2a.d[/tex]..............................(1)
where:
v=final velocity of the body
u=initial velocity of the body
here, since the body starts from rest state:
[tex]u=0m.s^{-1}[/tex]
putting the values in eq. (1)
[tex]v^2=0^2+2\times 3.62 \times 10[/tex]
[tex]v=8.5088\,m.s^{-1}[/tex]
Now, the momentum of the body just before the jump onto the tyre will be:
[tex]p=M.v[/tex]
[tex]p=20\times 8.5088[/tex]
[tex]p=170.1764\,kg.m.s^{-1}[/tex]
Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.
[tex](M+m)\times v'=p[/tex]
[tex](20+10)\times v'=170.1764[/tex]
[tex]v'=5.6725\,m.s^{-1}[/tex]
(b)
Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.
So,
[tex]\frac{1}{2} (M+m).v'^2=(M+m).g.h[/tex]
[tex]\frac{1}{2} (20+10)\times 5.6725^2=(20+10)\times 9.8\times h[/tex]
[tex]h\approx 1.6420\,m[/tex]
above the normal hanging position.
