A particle of charge +q and mass m is shot from rest at an angle theta with spee

Question

A particle of charge +q and mass m is shot from rest at an angle theta with speed initially unknown with respect to the horizontal direction. Let the origin of the coordinate system be the initial position of the charge. Let the horizontal direction be the x-axis and the vertical be the y-axis. Assume the only force on the charge is th eelectric field, E. A second mass M has no net charge and is floating in space with coordinates (d,h). The first mass is moving in the horizontal direction when it strikes the second mass and sticks to it thus the two become a single composite mass. The composite then moves according to Newton's second laws. In terms of the quantities given determine the following.

a) The acceleration of the first mass before it strikes the second mass.

b) the initial speed of the first mass

c) The speed of the composite mass immediately following the collision

d) The acceleration of the composite mass

e) The position on the x-axis where the composite mass crosses the x-axis

f) The speed of the composite mass when the composite mass crosses the x-axis

Explanation / Answer

a) F = q*E

==> a = q*E/m


b) let U is the initial speed

Ux = U*cos(theta)

Uy = U*sin(theta)


c)
m*U*cos(theta) = (m+M)*v

v = m*U*cos(theta) / (m+M)


d) a' = q*E/(m+M)


e)

let t is the time taken to fall on ground.

h = 0.5*a'*t^2

t = sqrt(2*h/a')

distance travelled after collision = t*u*cos(theta)

so at X = d + u*cos(theta)*sqrt(2*h/a') both bodies touch the ground.


f)Vy = sqrt(2*a'*h) = sqrt(2*q*E*h/(m+M) )

vx = m*u*cos(theta)/ (m+M)

v = sqrt(vx^2 + vy^2)

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