Simple and Physical Pendulums AP Physics C Mechanics FRQ – Exam Style Questions etc.
Simple and Physical Pendulums AP Physics C Mechanics FRQ
Unit 7: Oscillations
Weightage : 10-15%
Question
A small dart of mass 0.020 kg is launched at an angle of 30° above the horizontal with an initial speed of 10 m/s. At the moment it reaches the highest point in its path and is moving horizontally, it collides with and sticks to a wooden block of mass 0.10 kg that is suspended at the end of a massless string. The center of mass of the block is 1.2 m below the pivot point of the string. The block and dart then swing up until the string makes an angle θ with the vertical, as shown above. Air resistance is negligible.
(a) Determine the speed of the dart just before it strikes the block.
(b) Calculate the horizontal distance d between the launching point of the dart and a point on the floor directly below the block.
(c) Calculate the speed of the block just after the dart strikes.
(d) Calculate the angle θ through which the dart and block on the string will rise before coming momentarily to rest.
(e) The block then continues to swing as a simple pendulum. Calculate the time between when the dart collides with the block and when the block first returns to its original position.
(f) In a second experiment, a dart with more mass is launched at the same speed and angle. The dart collides with and sticks to the same wooden block.
i. Would the angle θ that the dart and block swing to increase, decrease, or stay the same?
_____ Increase _____ Decrease _____ Stay the same
Justify your answer.
ii. Would the period of oscillation after the collision increase, decrease, or stay the same?
_____ Increase _____ Decrease _____ Stay the same
Justify your answer.
Answer/Explanation
Ans:
(a)
Vtop = ?
Vtop, y = O m/s because top of protectory
Vtop,x only speed: Vo cosθ = 10 m/s (Cos 300) = 8,66 m/s
(b)
d = ? Vtop, y = O, V0, y = V0sin 300
d = Vx (t) \(\frac{V_{0}y}{g}=t\)
d = 8.66 m/s × 0.51s \(t = \frac{V_{0}sin30^{0}}{g}=0.51s\)
(c)
Vblock = ?
Cons. of angular momentum:
Li = Lf
rmv = Iw, \(w = \frac{v}{r}\) M = mdart + Mblock
\(rmv = MR^{2}(\frac{V}{R})\)
mdart (V0) = mdart + Mblock) (Vt)
10 m/s (OtoZ) = (0.oz + 0.1) (Vf)
Vf = 1.67 m/s
(d)
θ = ?
\(\frac{1}{2}mv^{2}=mgh\)
\(\frac{1}{2}(V_{f})^{2}=gh\) \(\frac{1}{2}(1.67 m/s)^{2}=9.81 m/s (h)\)
h = 0.14 m
\(cos\theta = \frac{a}{L}\)
\(cos\theta = \frac{1.2 m – h}{1.2 m}\)
\(cos\theta = \frac{1.2 – 0.14m}{1.2 m}\)
\(\theta = 28.2^{0}\)
(e)
Time = T/2
\(T = 2\pi \sqrt{\frac{L}{q}} \) simple pendulum
\(\frac{T}{2} = \pi \sqrt{\frac{L}{q}} \)
\(\frac{T}{2} = \pi \sqrt{\frac{1.2}{9.81}} \Rightarrow T/2 = 1.15\)
(f) i.
x Decrease
because the final velocity would be less and thus answer angle;
ii. x Stay the same
\(T = 2\pi \sqrt{\frac{L}{g} }\) the period would stay the same because the length and a stays the same despite change in mass