Newton’s Third Law AP Physics 1 FRQ – Exam Style Questions etc.
Newton’s Third Law AP Physics 1 FRQ
Unit: 2. Force and Translational Dynamics
Weightage : 10-15%
Exam Style Practice Questions, Newton's Third Law AP Physics 1 FRQ
Question: (7 points, suggested time 13 minutes)
Two blocks are connected by a string of negligible mass that passes over massless pulleys that turn with negligible friction, as shown in the figure above. The mass m2 of block 2 is greater than the mass m1 of block 1. The blocks are released from rest.
(a) The dots below represent the two blocks. Draw free-body diagrams showing and labeling the forces (not components) exerted on each block. Draw the relative lengths of all vectors to reflect the relative magnitudes of all the forces.
(b) Derive the magnitude of the acceleration of block 2. Express your answer in terms of m1 , m2 , and g.
Block 3 of mass m3 is added to the system, as shown below. There is no friction between block 3 and the table.
(c) Indicate whether the magnitude of the acceleration of block 2 is now larger, smaller, or the same as in the original two-block system. Explain how you arrived at your answer.
▶️Answer/Explanation
Ans:
(a) 
(b) According to Newton’s Second Law
For block 2: m2g – FT’ = m2.a
For block 1: Ft – m1g = m1.a.
According to Newton’s 3rd Law = |FT| = |FT’|
∴ m2g – m1g = m2a + m1a.
∴\(a^{2}=\frac{\left ( m_{2}-m_{1} \right )g}{m_{1}+m_{2}}\)
(c) a2 = a system = \(\frac{\left ( m_{2}-m_{1} \right )g}{m_{1}+m_{2}+m_{3}}< \frac{\left ( m_{2}-m_{1} \right )g}{m_{1}+m_{2}}=a_{2} original\)
∴ the acceleration now is smaller than that before due to the increase in the total mass of the system.
Question: (7 points, suggested time 13 minutes)
Two identical spheres are released from a device at time t = 0 from the same height H, as shown above. Sphere A has no initial velocity and falls straight down. Sphere B is given an initial horizontal velocity of magnitude v0 and travels a horizontal distance D before it reaches the ground. The spheres reach the ground at the same time tf , even though sphere B has more distance to cover before landing. Air resistance is negligible.
(a) The dots below represent spheres A and B. Draw a free-body diagram showing and labeling the forces (not components) exerted on each sphere at time \(\frac{t_{f}}{2}.\)
(b) On the axes below, sketch and label a graph of the horizontal component of the velocity of sphere A and of sphere B as a function of time.
(c) In a clear, coherent, paragraph-length response, explain why the spheres reach the ground at the same time even though they travel different distances. Include references to your answers to parts (a) and (b).
▶️Answer/Explanation
Ans:
(a)
(b)
(c)
Reaching the ground from the table concerns only vertical distance. Thus, the only component of velocity that impacts the time to reach the ground is the y component of velocity. The X-component makes a ball travel farther horizontally during the same time, but it won’t make it fall at a faster speed toward the ground. This can also be shown using the following Kinematic equation:
\(y = y_{0}+y_{g0}t.\frac{1}{2}a_{g}t^{2}\)
y final and y0 are the same for both spheres because they start and end at the same heights. The initial y component of velocity is zero for both spheres. Acceleration is also equal, for gravely is the only think acting on the spheres in the vertical direction. Thus, the time if takes for the spheres to reach the ground must be equal.