An electric cannon, such as shown in (Figure 1) , consists of a 10-kg metal ball

Question

An electric cannon, such as shown in (Figure 1) , consists of a 10-kg metal ball with charge 1.3×10?4 Ccompressed into a plastic barrel so that it is 0.10 m from an equally charged object at the closed end of the barrel. The barrel is oriented at 37? with respect to the horizontal. When the ball is released, it shoots 3.0 malong the barrel from its starting position because of the repulsive force between the two charged objects.

Q1. Find the change in gravitational potential energy.

Q2. Find the change in electric potential energy.

Q3. Find the final speed of the charge as it leaves the barrel.

Explanation / Answer

Q1) change in gravitational potential energy is U2-U1 = mg*(h2-h1)

h1 = 0.1*sin(37) = 0.06018 m
h2 = 3*sin(37) = 1.805 m

h2-h1 = 1.805-0.06018 = 1.745 m

U2-U1 = 10*9.8*(1.745) = 171.01 J

Q2) change in electrical potential energy is U2-U1 = -k*q1*q2[(1/r2)-(1/r1)]

(1/r2)-(1/r1) = (1/3)-(1/0.1) = -9.667

then U2-U1 = -9*10^9*1.3*1.3*10^-8*(-9.667) = 1470.35 J

Q3) From law of conseravation of energy

change in kinetic energy = chnage in potential energy

0.5*m*v^2 = 1470.35+171.01 = 1641.36 J

final speed is v = sqrt(1641.36/(0.5*10))

then v = 18.12 m/s

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