A metal cannonball of mass m rests next to a tree at the very edge of a cliff 36
ID: 2112129 • Letter: A
Question
A metal cannonball of mass m rests next to a tree at the very edge of a cliff 36.0 m above the surface of the ocean. In an effort to knock the cannonball off the cliff, some children tie one end of a rope around a stone of mass 80.0 kg and the other end to a tree limb just above the cannonball. They tighten the rope so that the stone just clears the ground and hangs next to the cannonball. The children manage to swing the stone back until it is held at rest 1.80 m above the ground. the children release the stone, which then swings down and makes a head-on, elastic collision with the cannonball, projecting it horizontally off the cliff. The cannonball lands in the ocean a horizontal distance R away from its initial position. (a) Find the horizontal component R of the cannonball's displacement as it depends on m. (b) What is the maximum possible value for R, and (c) to what value of m does it correspond? (d) For the stone-cannonball-Earth system, is mechanical energy conserved throughout the process? Is this principle sufficient to solve the entire problem? Explain. (e) What if? Show that R does not depend on the value of the gravitational acceleration. Is this result remarkable? State how one might make sense of it.
Explanation / Answer
By conservation of energy the velocity u at which the stone strikes the cannonball is-
u=sqrt(2gh)=sqrt(2g(1.8))=sqrt(3.6g)
Let veloctiy of the cannonball just after collision be v
In elastic collision-
velocity of seperation=velocity of approach , or
u=v-u1 ..........(equ1)
By conservation of linear momentum-
80(u)=80u1+mv
From equ1, u1=u+v
Thus,
80u=80(u+v)+mv
v=160u/(80+m)=160(sqrt(3.6g))/(80+m)
Total time taken by the cannonball to reach the ground = t considering the vertical velocity of the cannonball u =0
By 2nd kinemetical equation s=ut+g(t^2)/2
36=g(t^2)/2
t=sqrt(72/g)
R=v*t=[160(sqrt(3.6g))/(80+m)]*[sqrt(72/g)]...........(equ1)
Thus R=160(36)/[(80+m)sqrt(5)]=5760/[(80+m)sqrt(5)]=2575.95/(80+m)
=2576/(80+m)
a) R=2576/(80+m)
c) Thus max value of R is when m=0
b) R(max)=32.2 m
d) Since the collision between the stone and the cannonball is ellastic, and because the gravitational force is conservative force thus total energy is conserved.
However the principle is not sufficient to solve the entire problem as it gives us the data of only the total speed of the cannonball at any given time. We can't deduce the horizontal velocity during the projectile motion. Thus making it impossible to find the horizontal displacement R.
e)From equation 1 we can see that the sqrt(g) term gets cancelled making the R independent of g.
By this we can surely say that the horizontal distance covered the ball is independent of the vertical force acting on the cannonball. i.e. Horizontal distance R is not dependent on the magnitude of the vertical force acting on the cannonball(mg in this case)