A toy cannon uses a spring to project a 5.31 g soft rubber ball. The spring is o

Question

A toy cannon uses a spring to project a 5.31 g soft rubber ball. The spring is originallycompressed by 4.96 cm and has a forceconstant of 8.02 N/m. When the cannon isfired, the ball moves 14.3 cm through thehorizontal barrel of the cannon, and there is a constant frictionalforce of 0.0330 N between the barrel andthe ball. (a) With what speed does the projectile leavethe barrel of the cannon?
1 m/s

(b) At what point does the ball have maximum speed?
2 cm (from its original position)

(c) What is this maximum speed?
3 m/s (a) With what speed does the projectile leavethe barrel of the cannon?
1 m/s

(b) At what point does the ball have maximum speed?
2 cm (from its original position)

(c) What is this maximum speed?
3 m/s

Explanation / Answer

Let speed with which bullet leave the barrel be v m/s Initail Energy - Final Energy = Workdone againstfriction Initial energy = potential energy of spring = (1/2)Kx2 = (1/2)(8.02 N/m)(0.0496 m)2 = 0.009865N-m Final energy = kinetic energy of bullet = (1/2)mv2= (1/2)(0.00531 kg)(v m/s)2 = 0.002655v2N-m Work done against friction = FrL = (0.0330 N)(0.143 m) = 0.004719 N-m Equating , 0.009865 N-m - 0.002655v2 N-m = 0.004719N-m v = 1.3922 m/s ...........(a) Initially (for first 4.96cm) spring will be accelarating thebullet and friction will be slowing it down. After that frictionwill be slowing the bullet and there is no acceleration. So, maximum velocity will be at the point which is4.96 cm from the original position ............(b) Doing the same energy calculations, but now at thispoint. Initial energy will remain same = 0.009865 N-m Final energy = (1/2)(m)(vmax)2 =0.002655(vmax)2 N-m workdone against friction = FrL = (0.0330 N)(0.0496 m) =0.001637 N-m Equating and solving for vmax vmax = 1.7604 m/s.........(c) vmax = 1.7604 m/s.........(c)

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