Respuesta :
Answer:
Final speed is 8.506 m/s
Height of the block is 0.1 m
Solution:
As per the question:
Mass of the smaller block, m = 5 kg
Height of the track, h = 5 m
Mass of the larger block, M = 18 kg
Now,
To calculate the final speed of the larger block:
Consider that both the momentum and the energy of the block is conserved as the track is friction-less.
If the smaller block is at a height 'h' then it will have potential energy only.
The initial energy of the entire system is ;
[tex]E_{i} = PE = mgh[/tex]
As it starts moving on the track, the potential energy is completely transfomed into Kinetic energy that provides motion, thus the final enrgy:
[tex]E_{f} = KE = \frac{1}{2}mv^{2}[/tex]
Following the conservation of energy:
[tex]E_{i} = E_{f}[/tex]
[tex]mgh = \frac{1}{2}mv^{2}[/tex]
[tex]v = \sqrt{2gh}[/tex]
Since, the collision is elastic, both Kinetic energy of the blocks and the momentum are conserved.
Now,
Initial Momentum, [tex]p_{initial} = mv[/tex]
Final Momentum, [tex]p_{final} = -mv' + Mv''[/tex]
Now, by the conservation of momentum:
[tex]p_{initial} = p_{final}[/tex]
[tex]mv = -mv' + Mv''[/tex]
[tex]m(v + v') = Mv''[/tex] (1)
From the conservation of kinetic energy:
[tex]\frac{1}{2}mv^{2} = \frac{1}{2}mv'^{2} + \frac{M}{v''^{2}}[/tex]
[tex]m(v^{2} - v'^{2}) = Mv''^{2}[/tex] (2)
Dividing eqn (1) and (2), we get:
v = v' + v''
Also, from above, we can write:
v' + v'' = [tex]\sqrt{2gh}[/tex]
v' + v'' = [tex]\sqrt{2gh}[/tex]
v' = [tex]\sqrt{2gh}[/tex] - v'' (3)
Using eqn (3) in eqn (1) and (2) and solving we get:
[tex]\sqrt{2gh}\frac{2m}{M + m}[/tex]
Using proper values in the above eqn:
[tex]v'' = \sqrt{2\times 9.8\times }\frac{2\times 5}{18 + 5} = 8.506\ m/s[/tex]
[tex]v' = \sqrt{2\times 9.8\times 5} - 8.506 = 1.393\ m/s[/tex]
Now, the Kinetic energy required to climb the curve:
[tex]E_{i} = \frac{1}{2}mv'^{2}[/tex]
Now, at height h':
[tex]E_{f} = mgh'[/tex]
By the conservation of energy principle:
[tex]\frac{1}{2}mv'^{2} = mgh'[/tex]
[tex]h' = \frac{v^{2}}{2g}[/tex]
[tex]h' = \frac{1.393^{2}}{2\times 9.8} = 0.1 m[/tex]
