Question

5. The ball launcher in a pinball machine has a spring that has a force constant of 1.20 N/cm . The surface on which the ball moves is inclined 10.0° with respect to the horizontal. The spring is initially compressed 5.00 cm. Find (a) the launching speed of a 100-g ball when the plunger is released; (b) How far does the ball move along the incline? Friction and the mass of the plunger are negligible.

Answer 1

a. Wb = m*g = 0.1kg * 9.8N/kg = 0.98 N. = Wt of the ball.

Fb = 0.98N @ 10o = Force of ball.
Fp = 0.98*sin10o = 0.170 N. = Force parallel to incline.
Fv = 0.98*cos10 = 0.965 N. = Force perpendicular to incline.

Fap = 1.2N/cm * (-5cm) = -6 N, = Force applied.

a=Fn/m=(Fap-Fp)/m
a=(-6+0.17)/0.1=-58.3m/s^2

Ekmax = Epmax.
0.5m*V^2 = mg*d.
0.5*0.1*V^2 = 0.98*5.
05*V^2 = = 4.9
V^2 = 98
V = 9.9 m/s.

b. V^2 = Vo^2 + 2a*d.
d = (V^2-Vo^2)/2a=(0-98)/-116.6=0.84 m.