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kondaur [170]
3 years ago
7

projectile motion of a particle of mass M with charge Q is projected with an initial speed V in a driection opposite to a unifor

m electric fiedl of magnitude E.
Physics
1 answer:
SIZIF [17.4K]3 years ago
5 0

Answer:

Range, R = MV²/2QE

Explanation:

The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.

Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses  all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.

So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops

Therefore {tex}R = MV²/2QE{/tex}

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3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
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4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
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And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36  m/s
with negative sign, because it is directed downward.
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