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

Answer:

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

b) $x=0.0451m$

c) $v_m=1.75m/s$

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

$\frac{1}{2}kc^2-f\triangle x=\frac{1}{2} mv^2$

$v=\sqrt{\frac{k*d_c^2-2f\triangle x}{m}}$

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

$v=1.102122995 \approx 1.10m/s$

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

$\frac{1}{2}kc^2-f\triangle x'=\frac{1}{2} mv'^2+\frac{1}{2}kc'^2$

$\frac{1}{2} mv'^2 = \frac{1}{2}kc^2-f\triangle x'-\frac{1}{2}kc'^2$

$v=\sqrt{\frac{k*d_c^2-2*f*x-kd_c'^2}{m}}$

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

$v=1.749685197m/s$

$v_m=1.75m/s$