A particle is thrown upward with speed 10√3 m/sec. It strikes an inclined surface. Collision between the particle and inclined plane is perfectly inelastic. What will be the maximum height attained by the particle from the ground.
2. Relevant information and equations
- In perfectly inelastic collision (e=0) colliding objects stick to each other.
- When net F external = 0, momentum is conserved.
- v2 = u2 + 2as (for constant acceleration a)
The Attempt at a Solution
Just before collision let velocity of particle be v
⇒ v2 = (10√3)2 - 2g(3)
⇒ v = 4√15 m/s (g=10 m/s2
Let m = mass of particle, M = mass of surface, V = velocity of particle and surface along the normal.
Along the normal,
∫Ndt = m(V-vcos45°) = MV
⇒V = mv/√2(m-M)
Since denominator is very large, V≈ 0
Now component of velocity along the inclined plane ( = vsin45°) will not change as the normal reaction (which will give the impulse during collision) is acting to perpendicular to it.
So finally particle will move along the incline particle( upwards) till its velocity becomes zero.
0 = (vsin45°)2 - 2s(gsin45°)
⇒ s = 6√2
Thus, total height above ground = s(sin45°) + 3 = 9 m
But the answer is 6 m.
Is it because I neglected the velocity along the normal?