An electric cannon, such as shown in (Figure 1) , consists of a 20-kg metal ball

Question

An electric cannon, such as shown in (Figure 1) , consists of a 20-kg metal ball with charge 1.2×104 C compressed 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 m along the barrel from its starting position because of the repulsive force between the two charged objects.

Part A: Find the change in gravitational potential energy.

Part B: Find the change in electric potential energy.

Part C: Find the final speed of the charge as it leaves the barrel.

Explanation / Answer

Part A:

Change in gravitational potential energy of the cannon = Gravitational potential energy attained due to its height by the cannon

Now, height attained by the cannon, h = d * sin37 = 3.0 * 0.602 = 1.805 m

So, change in gravitational potential energy = m * g * h

= 20 * 9.81 * 1.805 = 354.14 J

Part B: Apply conservation of energy -

change in gravitational potential energy = change in electric potential enery = 354.14 J

Part C:

To find the final speed, here also apply conservation of energy.

Change in potential energy = change in kinetic energy

=> m*g*h = (1/2)*m*v^2

=> v^2 = 2*g*h

=> v = sqrt[2*g*h] = sqrt[2*9.81*1.805] = 5.95 m/s.

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