How Does Friction Affect the Speed of a Toy Cannon Ball?

[wachaif]

A toy cannon uses a spring tp project a 5.3 g soft rubber ball. The spring is originally compressed by 5 cm and has a force constant 8N/m .When the cannon is fired , the ball moves 15cm through the horizontal barrel of the cannon ,and there is a constant friction force of 0.032 N . between the barrel and the ball.At what point does the ball have maximum speed ? What is this maximum speed?

 

[HallsofIvy]

The force due to the spring at any instant is the spring constant times the compression at that instant: If we take x= 0 at the point the ball is when the spring is compressed, then the force due to the spring is 8(0.05-x) (as long as x< 0.05 m). The friction force is 0.032 (as long as x< 0.15 cm) so the total force on the ball is:
8(0.05-x)- 0.32= 0.08- 8x for x< 0.05
= -0.032 for 0.05< x< 0.15

Since "force= mass* acceleration", we have the differential equation

0.0053x"= 0.08- 8x for x< 0.05
= -0.032 for 0.05< x< 0.15
with initial conditions x(0)= 0, x'(0)= 0.

 

[M98Ranger]

The ball will have maximum speed when it is released from the spring, as this is when it has the most potential energy that can be converted into kinetic energy. This occurs at the end of the 5 cm compression, right before it begins to move through the barrel.

To calculate the maximum speed, we can use the equation for potential energy (PE) and kinetic energy (KE):

PE = KE

1/2 kx^2 = 1/2 mv^2

Where k is the force constant, x is the distance the spring is compressed, m is the mass of the ball, and v is the velocity.

Plugging in the given values, we get:

1/2 (8N/m)(0.05m)^2 = 1/2 (0.0053kg)v^2

Solving for v, we get a maximum speed of 2.07 m/s.

This means that at the end of the 5 cm compression, the ball will have a speed of 2.07 m/s before it begins to decelerate due to the friction force.