Question

A toy cannon uses a spring to project a 5.26 g soft rubber ball. The spring is originally compressed by 4.93 cm and has a stiffness constant of 7.90 N/m. When it is fired, the ball moves 14.6 cm through the horizontal barrel of the cannon, and there is a constant frictional force of 0.0328 N between barrel and 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 and process please

Answer 1

a)
`v=1.102122995 \approx 1.10m/s`

b)
`x=0.0451m`

c)
`v_m=1.75m/s`

Step-by-step explanation:

From the question we are told that

Mass of soft rubber
`M_r= 5.26g=>0.00526kg`

Spring compression
`d_C= 4.93 cm=>0.0493m`

Stiffness constant
`k= 7.90 N/m`

Ball Distance
`d=14.6=>0.146`

Frictional force
`f=0.0328 N`

a)Generally the equation for motion of the second ball according to Newton's law is given by

`(1)/(2)kc^2-f\triangle x=(1)/(2) mv^2`

`v=\sqrt{(k*d_c^2-2f\triangle x)/(m)}`

`v=\sqrt{((7.90 )(0.0493)^2-2*(0.0328)(0.0493+0.146))/(0.00526)}`

`v=1.102122995 \approx 1.10m/s`

b)Generally for speed to be at maximum spring force must equal frictional force

At distance

`f=kc'`

`d_c'=f/k`

`d_c'=0.0328 /7.90`

`d_c'=0.004151898734=>0.0042m`

Therefore

Ball is at maximum speed at

`x=d_c-d_c'`

`x=0.0493-0.0042`

`x=0.0451m`

c)Generally maximum velocity is represented mathematically as

`(1)/(2)kc^2-f\triangle x'=(1)/(2) mv'^2+(1)/(2)kc'^2`

`(1)/(2) mv'^2 = (1)/(2)kc^2-f\triangle x'-(1)/(2)kc'^2\\`

`v=\sqrt{(k*d_c^2-2*f*x-kd_c'^2)/(m)}`

`v=\sqrt{(7.90 *0.0493^2-2*(0.0328)*(0.0451)-7.90(0.0042)^2)/(0.00526)}`

`v=1.749685197m/s`

`v_m=1.75m/s`