Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How f
1 answer:
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Answer:
The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.
Explanation:
Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively
and the mass of big rock be 'M'
Initial momentum of the system equals
Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'
Thus the final momentum of the system is
Equating initial and the final momenta we get
Now since the surface is frictionless thus the energy is also conserved thus
Similarly the final energy becomes
\
Equating initial and final energies we get
Solving i and ii we get
Using this in equation i we get
Thus putting v = -v' in equation i we get V' = 0
This implies Smaller stone rebounds while as larger stone remains stationary.
Answer:
0.331m
Explanation:
wave equation: v=f λ
v=30000m/s
f=90500 Hz
λ=
λ=
λ= 0.311m