E = MC^2. Albert Einstein's proven formula. When mass travels at the square of speed of light, the mass gets converted into energy
We will measure all angles from West, the negative x-axis and divide the journey into 3 parts:
P1 = 370y
P2 = 410cos(45)x + 410sin(45)y = 290x + 290y
P3 = 370cos(270 - 28)x + 370sin(270 - 28) = -174x - 327y
Overall displacement:
x = 290 - 174 = 116 m
y = 370 + 290 - 327 = 333 m
displacement = √(116² + 333²)
= 353 m
Direction:
tan(∅) = y/x
∅ = tan⁻¹ (333 / 116)
∅ = 70.8° from West.
Answer:

Explanation:
GIVEN
diameter = 15 fm =
m
we use here energy conservation

there will be some initial kinetic energy but after collision kinetic energy will zero

on solving these equations we get kinetic energy initial
J ..............(i)
That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e
and gains kinetic energy K =e∆V ..........(ii)
by accelerating through a potential difference ∆V
Thus the alpha particle will
just reach the
nucleus after being accelerated through a potential difference ∆V
equating (i) and second equation we get
e∆V = 35.33 Me V
