Elastic and Inelastic Collisions AP Physics C Mechanics FRQ – Exam Style Questions etc.
Elastic and Inelastic Collisions AP Physics C Mechanics FRQ
Unit 4: Linear Momentum
Weightage : 15-25%
Question
A pendulum of length L consists of block 1 of mass 3M attached to the end of a string. Block 1 is released from rest with the string horizontal, as shown above. At the bottom of its swing, block 1 collides with block 2 of mass M, which is initially at rest at the edge of a table of height 2L. Block 1 never touches the table. As a result of the collision, block 2 is launched horizontally from the table, landing on the floor a distance 4L from the base of the table. After the collision, block 1 continues forward and swings up. At its highest point, the string makes an angle θMAX to the vertical. Air resistance and friction are negligible. Express all algebraic answers in terms of M, L, and physical constants, as appropriate.
(a) Determine the speed of block 1 at the bottom of its swing just before it makes contact with block 2.
(b) On the dot below, which represents block 1, draw and label the forces (not components) that act on block 1 just before it makes contact with block 2. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot. Forces with greater magnitude should be represented by longer vectors.
(c) Derive an expression for the tension FT in the string when the string is vertical just before block 1 makes contact with block 2. If you need to draw anything other than what you have shown in part (b) to assist in your solution, use the space below. Do NOT add anything to the figure in part (b). For parts (d)–(g), the value for the length of the pendulum is L = 75 cm.
(d) Calculate the time between the instant block 2 leaves the table and the instant it first contacts the floor.
(e) Calculate the speed of block 2 as it leaves the table.
(f) Calculate the speed of block 1 just after it collides with block 2.
(g) Calculate the angle θMAX that the string makes with the vertical, as shown in the original figure, when block 1 is at its highest point after the collision.
Answer/Explanation
Ans:
(a)
\(mgh = \frac{1}{2}mv^{2}\)
10 (L) = .sv2 v2 = 20L
\(V=\sqrt{20L}=2\sqrt{SL}\)
(b)
(c)
FT = ma
FT – Fg = 3M.
(d)
Xy= Xoy + Voy t+ 1/2 at2
-2L = 0 + 0 t+ 1/2 (-10) t2
-1.5 m = -5t2
3 = t2
\(t = \sqrt{\frac{3}{10}}=0.548 seconds\)
(e)
\(V = \frac{\Delta x}{\Delta t}= \frac{4L}{\sqrt{\frac{3}{10}}}=\frac{3}{\sqrt{\frac{3}{10}}}=5.477 m/s\)
(f)
P1 + P2 = P1f + P2f
m1v1 + m2v2 = m1v1f + m2v2f
\((3m)\left ( \sqrt{20(.70)} \right )+(m)(0)=(3m)(x)+(m)(5.477)\)
11.619 + 0 = 3x + 5.477
6.142 = 3x
x = 2.047 m/s
(g)
\(\frac{1}{2}3MV^{2}=3Mgh\)
\(\frac{1}{2}(2.047)^{2}=10 h\)
H = 0.210 m
\(cos(\theta )=\frac{A}{H}\)
\(\theta = -cos^{-1}\left ( \frac{A}{H} \right )\)
\(\theta = cos^{-1}\left ( \frac{.75-H}{.75} \right )=43.897^{0}\)