Solution :
Given :
M = 0.35 kg

Total mechanical energy = constant
or 
But
and 
Therefore, potential energy at the top = kinetic energy at the bottom


(h = 35 cm = 0.35 m)
= 2.62 m/s
It is the velocity of M just before collision of 'm' at the bottom.
We know that in elastic collision velocity after collision is given by :

here, 
∴ 

= 0.33 m/s
Therefore, velocity after the collision of mass M = 0.33 m/s
Acceleration occurs when velocities change.velocity changes either because the speed changes or the direction change.
Hard surfaces reflect sound back into the room, while carpets help to absorb the sound so it reflects less
Answer:
The kinetic energy of the particle as it moves through point B is 7.9 J.
Explanation:
The kinetic energy of the particle is:
<u>Where</u>:
K: is the kinetic energy
: is the potential energy
q: is the particle's charge = 0.8 mC
ΔV: is the electric potential = 1.5 kV
Now, the kinetic energy of the particle as it moves through point B is:


Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.
I hope it helps you!