Answer to Question #145606 in Mechanics | Relativity for Marah

Question

A toy cannon uses a spring to project a 5.38-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force constant of 8.05 N/m. When the cannon is fired, the ball moves 14.9 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.032 2 N on the ball.
(a) With what speed does the projectile leave the barrel of the cannon?
(b) At what point does the ball have maximum speed? cm (from its original position)
(c) What is this maximum speed?

Answer 1

a) energy gained from the spring is,

$1/2kx^{2} = 1/2 \times 8.05\times0.0508^{2}=0.0104J$

energy lost due to due to friction = $F_{f}\times d=0.0322\times 0.149=0.0048J$

resultant energy = $0.0104J - 0.0048J=0.0056J$

$1/2mv^{2} = 0.0056$

$0.5\times5.38\times10^{-3}\times v^{2}=0.0056$

$v^{2}=2.0818$

$v=\sqrt{2.0818}=1.44m/s$

(b) point of maximum speed$=0.0508-\frac{0.0322}{8.05}=0.0508-0.004=0.0468m$

(c) $E_{gain}= 0.010-1/2\times8.05\times0.004^{2}=0.010-0.0000644=0.00994J$

$E_{energy}$ lost due to friction = $0.0322 \times 0.0468 = 0.0015J$

resultant energy =$0.00994-0.0015=0.00844J$

$1/2mv^{2}=0.00844$

$0.5\times 5.38\times10^{-3}\times v^{2}=0.00844$

$v^{2} = \frac{0.00844}{0.5\times 5.38 \times10^{-3}}=3.1375$

$v = 1.77m/s$