1.A steel ball (66.45 g), shot in the horizontal direction, is caught by a stati

Question

1.A steel ball (66.45 g), shot in the horizontal direction, is caught by a stationary pendulum (275.05 g). The pendulum rises 8.55 cm in the vertical direction. Calculate the momentum of the ball-pendulum system at the moment the ball is caught by the pendulum.

2.A steel ball (66.45g) located 103.15 cm above the ground is shot in the horizontal direction. It travels 306.25 cm in the horizontal direction before it hits the ground. Calculate the momentum of the steel ball in the horizontal direction. Enter your results using the measured units, i.e., grams, centimeters and seconds. Use scientific notation for this calculation.

Calculate the percent fractional error of the loss of momentum to two significant figures.(PFE =|pf - pi/pi|*100)

Explanation / Answer

1) at first we have to find the initial velocity (v) of the pendulum+ball by conservation of energy ..
Grav.PE gained at max. height = initial KE .. (assuming no other energy transfers occur)
Mgh = ½ Mv²
v = (2gh) = (2*9.8*8.55^-2m) = 1.29 m/s

Initial mom.(joint mass) = Mv = (0.275+0.0665)kg * 1.29m/s initial momentum = 0.44 kg.m/s

2) 103.15 cm = 1.0315 m

66.45 g = 0.06645 kg

s = ut + ½at²

vertically u = 0 = initial velocity

=> 1.0315 = ½9.81t²

=> t = [2.0630/9.81] = 0.46 s

horizontally the ball's speed is constant since unaffected with vertical g


=> distance = speed x time

=> horizontal speed = distance/time = 3.0625 / 0.46 s ~= 6.678 m/s = 667.8 cm/s

momentum = mv = 0.06645 x 6.678 ~= 0.444 Ns or 0.444 kgm/s

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