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 1.6×104 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.

Part A

What quantities can be determined using this information?

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Correct

Part B

Find the change in gravitational potential energy.

Express your answer to two significant figures and include the appropriate units.

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Correct

Part C

Find the change in electric potential energy.

Express your answer to two significant figures and include the appropriate units.

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Part D

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

Express your answer to two significant figures and include the appropriate units.

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the change in electric potential energy the change in gravitational potential energy size of metal ball the final speed of the charge as it leaves the barrel

Explanation / Answer

Here ,

mass of ball , m = 10 Kg

charge , q = 1.6 *10^-4 C

theta = 37 degree

part a)

the quantities that can determined

the change in electric potential energy
the change in gravitational potential energy
the final speed of the charge as it leaves the barrel

part b)

change in gravitational potential energy = m * g * sin(theta) * (L)

change in gravitational potential energy = 10 * 9.8 * sin(37) * 3

change in gravitational potential energy = 177 J

the change in gravitational potential energy is 177 J

part c)

change in electric potential energy = k * q * q * (1/d2 - 1/d1)

change in electric potential energy = 9*10^9 * 1.6 *10^-4 * 1.6 *10^-4 *(1/3.1 - 1/.1)

change in electric potential energy = -2229.7 J

the change in electric potential energy is -2229.7 J

part d)

let the final speed is v

using conservation of energy

0.5 * 10 * v^2 + 177 - 2229 = 0

solving for v

v = 20.3 m/s

the speed of the charge is 20.3 m/s

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