Where are you stuck? If you can answer the first question, you can easily answer the others too.
A particle of mass m is projected vertically upwards with speed u and when it reaches its greatest height a second particle of mass 2m is projected vertically upward with speed 2u from the same point as the first.
prove that the time elapses between the projection of the second particle and its collision with the first is u/4g
find the height above the ground at which this occurs
if on collision thwe particles coalesce, prove that the combined particle will reach a greates height of 19u^2/18g
many thanks
The u and 2u velocities are assumed to be vertical only.
When the 1m particle reaches its maximum height:
The rising of the particle stops, so effective vertical velocity becomes zero
u -g(t1) = 0 ..............read t1 as "t sub 1"
u = g(t1)
So, t1 = u/g
Then at time t1, the 2m particle is projected upwards.
The 1m particle starts to go down.
The 1m particle and the 2m particle will meet somewhere above the ground, say at H above the ground.
The 2m particle will need time t2 to get to H.
H = 2u(t2) -(g/2)(t2)^2 ------------------(i)
The 1m particle will need also t2 to reach the H from above,
Since the 1m particle starts fron zero velocity in its falling down, then the distance travelled by the 1m particle is
h2 = 0(t2) -(g/2)(t2)^2
h2 = -(g/2)(t2)^2
This will give a negative distance. Its positive or absolute distance is
h2 = (g/2)(t2)^2 ---------------------(ii)
H +h2 = h1 ------------------------(iii)
where h1 = maximum height reached by the 1m particle.
h1 = u(t1) -(g/2)(t1)^2
Substitute u/g for t1,
h1 = u(u/g) -(g/2)(u/g)^2
h1 = (u^2)/(2g) ------------------------(iv)
Hence, using (i), (ii) and (iv) into (iii),
2u(t2) -(g/2)(t2)^2 +(g/2)(t2)^2 = (u^2)/(2g)
2u(t2) = (u^2)/(2g)
t2 = [(u^2)/(2g)] / (2u)
t2 = u/(4g) -------------------------proven.
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find the height above the ground at which this occurs
That is H.
H = 2u(t2) -(g/2)(t2)^2 ------------------(i)
H = 2u[u/(4g)] -(g/2)[u/(4g)]^2
H = (u^2)/(2g) -(u^2)/(32g)
H = (15u^2) / (32g) --------------------------answer.
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if on collision the particles coalesce, prove that the combined particle will reach a greates height of 19u^2/18g
At collision, the 1m particle is going down and the 2m particle is going up. If they coalesce, the resulting single particle will be (1m +2m) = 3m in mass.
Here, as always, the momentum of the system will be conserved before and after the collision, or,
(m1)(v1) +(m2)(v2) = (m3)(v3) --------------(v)
So we find the
v1 = velocity of 1m particle at collision
v2 = velocity of 2m particle at collision
v3 = velocity of 3m particle after collision
v1 = 0 -g(t2) = -g(u / 4g) = -u/4
v2 = 2u -g(t2) = 2u -g(u / 4g) = 2u -u/4 = 7u/4
Substitutions into (v),
(1m)(-u/4) +(2m)(7u/4) = (3m)(v3)
-u/4 +14u/4 = 3(v3)
v3 = (13u/4) /3
v3 = 13u /12 -------------**
So the 3m particle has a positive velocity. It will go up some more, starting from the H above the ground.
Let us say it will go maximum of Y above the H
And it will take time t3 for it to do that.
At its highest, v3 = g(t3), so,
13u /12 = g(t3)
t3 = (13u / 12g)
Y = (v3)(t3) -(g/2)(t3)^2
Y = (13u /12)(13u / 12g) -(g/2)(13u / 12g)^2
Y = [169u^2 / 144g] -[169u^2 / 288g]
Y = 169u^2 / 288g
Hence,
H +Y = [15u^2 / 32g] +[169u^2 / 288g] = [(135u^2 +169u^2) / 288g] = 304u^2 / 288g = 19u^2 / 18g
Proven.