Please Read Question Below!

Question

A toy cannon uses a spring to project a 5.27-g soft rubber ball. The spring is originally compressed by 4.95 cm and has a force constant of 7.99 N/m. When the cannon is fired, the ball moves 16.0 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.0322 N on the ball.

(a) With what speed does the projectile leave the barrel of the cannon?

_________m/s

(b) At what point does the ball have maximum speed?

_________cm (from its original position)

(c) What is this maximum speed?

_________m/s

Answer 1

Hi Jeremy,

To solve this problem, we have to break the scenario down into 2 parts:

1. After the spring has been fully relaxed and the ball is no longer on the spring (I'm assuming no friction in this section),
2. The ball traveling down the barrel of the cannon.

(a) We must find the velocity of the ball after the spring is fully relaxed. Notice that the compressed spring has potential energy, and when the spring is relaxed, all of this energy has been transferred to the ball in the form of kinetic energy. This is nothing but Conservation of Energy. The initial potential energy of the spring is U = 1/2 * k * x², where k = 7.99 N/m is the force constant, and x = 4.95 cm = 0.0495 m is the compression of the spring. Plugging these numbers in gives U = 1/2 * (7.99 N/m) * (0.0495 m)² = 0.00979 J. Because of conservation of energy, this is also the kinetic energy of the ball.

For the second part of the ball's movement, it encounters a frictional force. The work done by the frictional force is equal to W = Fd, where F = -0.0322 N is the frictional force (the negative sign is there because is slows down the ball), and d = 16 cm = 0.16 m is the distance the ball moves. The work done by friction is then (-0.0322 N)(0.16 m) = -0.00515 J. The kinetic energy of the ball after it exits the barrel is

0.00979 J (the initial energy) - 0.00515 J (work done by friction) = 0.00464 J.

Kinetic energy of the ball is K = 1/2 * m * v² = 0.00464 J, where m = 5.27 g = 0.00527 kg . Solving for velocity v, we get v = √(2 * 0.00464 J / 0.00537 kg) = 1.31 m/s when it exits the barrel.

(b) The ball is at its fastest when it leaves the spring, as friction slows it down after that.

(c) From part (a), the kinetic energy of the ball when it leaves the spring is 0.00979 J, so the speed of the ball at this point is

v = √(2 * K / m) = √(2 * 0.00979 J / 0.00537 kg) = 1.91 m/s.

Hope this helps! Please reach out if you have questions.

Akshat Yaparla