A student drops a 1.5 kg rock off of a 4.0 m tall bridge. The speed of the rock just before it hits the water below is 6 m/s. Ho
2 answers:
You might be interested in
The magnetic force acting on a charged particle moving perpendicular to the field is:
= qvB
is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.
The centripetal force acting on a particle moving in a circular path is:
= mv²/r
is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.
If the magnetic force is acting as the centripetal force, set equal to
and solve for v:
qvB = mv²/r
v = qBr/m
Due to the work-energy theorem, the work done on the proton by the potential difference V becomes the proton's kinetic energy:
W = KE
W is work, KE is kinetic energy
W = Vq
KE = 0.5mv²
Therefore:
Vq = 0.5mv²
Substitute v = qBr/m and solve for V:
V = 0.5qB²r²/m
Given values:
m = 1.67×10⁻²⁷kg (proton mass)
B = 0.750T
q = 1.60×10⁻¹⁹C (proton charge)
r = 1.84×10⁻²m
Plug in the values and solve for V:
V = (0.5)(1.60×10⁻¹⁹)(0.750)²(1.84×10⁻²)²/1.67×10⁻²⁷
V = 9120V