Spring Forces AP Physics C Mechanics MCQ – Exam Style Questions etc.
Spring Forces AP Physics C Mechanics MCQ
Unit 2: Force and Translational Dynamics
Weightage : 20-15%
Question
Two blocks on a surface of negligible friction are attached together by a spring, as shown in the figure above. The mass of block 1 is M, and the mass of block 2 is 2M. Block 1 is initially moving toward block 2 , which is at rest. The spring compresses and decompresses in a repeated motion as the two blocks slide to the right. Which of the following statements best describes the speed of the center of mass of the two-block system?
A It decreases when the spring is compressed.
B It decreases when the spring expands.
C It is zero when the spring is fully compressed.
D It is at a maximum when the spring is stretched to a maximum.
E It is constant throughout the motion.
Answer/Explanation
Ans:E
The force of the spring is internal to the two blocks, so it cannot change the speed of the center of mass of the two-block system.Question
Two blocks of masses M and 2M are on a frictionless horizontal surface, as shown above, and are held in place with a compressed spring of negligible mass between them. If the blocks are then released and the block of mass 2M leaves the spring with a velocity v, the velocity of the center of mass of the blocks is
A zero\
B \(−\frac{v}{2}\)
C \(−\frac{2v}{3}\)
D \(−\frac{3v}{2}\)
E \(−2v\)
Answer/Explanation
Ans:A
Question
A 1-kg block hangs from a string at rest. A person holds a spring scale attached vertically to the string, pulling upward so that the scale reads 10 N. What is the tension in the string?
(A) 19 N
(B) 9 N
(C) 0 N
(D) 10 N
(E) 20 N
Answer/Explanation
Ans:
D—The block is in equilibrium, at rest. Thus the weight equals the tension, which is 10 N. The scale simply reads this tension. (If you wanted to do a free body diagram of the scale, it would include a 10 N tension pulling down, and the 10 N force of the hand pulling up.)
Question
The drag force on a falling coffee filter can be modeled as a linear function with respect to velocity, \(\vec{F}_D=−bv\), where b is a positive constant related to the surface area and shape of the coffee filter. The teacher demonstrates how the value of constant b can be determined by releasing coffee filters from rest and having them fall toward a motion detector, as shown. The students repeat the activity and use the motion detector to measure the position, velocity, and acceleration of different coffee filters as they fall. Which of the following would provide the data needed to determine the value of b?
A Cut coffee filters into different shapes of the same surface area and measure the terminal velocity of the coffee filters.
B Vary the diameter of the coffee filter and measure the final velocity of the coffee filter.
C Vary the starting height of the coffee filter and measure the final velocity of the coffee filter.
D Stack multiple coffee filters in order to vary the weight of the system without changing the surface area and measure the acceleration of the coffee filters.
E Stack multiple coffee filters in order to vary the weight of the system without changing the surface area and measure the terminal velocity of the coffee filters.
Answer/Explanation
Ans:E
At terminal velocity, the acceleration of the filters is zero. Applying Newton’s second law yields \(∑F=mg−bv=0\) ; therefore, the terminal velocity is proportional to the mass of the coffee filters. Using the equation \(mg=bv\)\(v=\frac{g}{b}m\), the slope of a graph of the terminal velocity as a function of mass can be used to calculate a value for b.
Question
An experiment is set up using a hanging mass to pull a cart along a horizontal track. A light string is attached from the hanging mass over a pulley to a force sensor. A second light string is attached from the force sensor to the cart. There is a motion sensor that can be used to measure different features of the motion of the cart. The pulley is ideal, and there is negligible friction between the cart and the track.
The cart of mass M starts at rest. The reading on the force sensor is F. Which of the following is a correct expression for the distance that the cart moved in time t ?
A \(\frac{Mt}{F}\)
B \(\frac{Ft}{M}\)
C \(\frac{Ft^2}{M}\)
D \(\frac{Ft^2}{2M}\)
E \(\frac{Mt^2}{2F}\)
Answer/Explanation
Ans:D
Applying Newton’s second law to the cart yields \(a=\frac{F}{M}\) . Substituting this acceleration into a kinematics equation for distance moved yields \(d=v_1t+\frac{1}{2}at^2 \rightarrow (0)t+\frac{1}{2}(\frac{F}{M})t^2=\frac{Ft^2}{2M}\)