Unit: 2. Force and Translational Dynamics
Weightage : 10-15%
An object of mass \(M\) is dropped near the surface of Earth such that the gravitational field provides a constant downward force on the object.
Which of the following describes what happens to the center of mass of the object-Earth system as the object falls downward toward Earth?
Correct Answer: \( \boxed{\mathrm{C}} \)
Consider the combined system consisting of the object and Earth.
The gravitational force that Earth exerts on the object and the gravitational force that the object exerts on Earth form an action-reaction pair.
These forces are internal to the object-Earth system.
Therefore, the net external force acting on the system is
\( \sum F_{\text{ext}} = 0 \)
For any system,
\( \sum F_{\text{ext}} = M_{\text{system}} a_{\text{CM}} \)
Since the net external force is zero,
\( a_{\text{CM}} = 0 \)
Thus, the center of mass of the object-Earth system does not accelerate.
As the object falls toward Earth, Earth also moves slightly toward the object. Their motions are such that the center of mass remains fixed.
Therefore, the center of mass of the object-Earth system
\( \boxed{\text{does not move}} \).
This result is independent of the values of Earth’s mass and the object’s mass, provided no external forces act on the system.
Three blocks are sliding together to the right along a surface of negligible friction when a force with magnitude \(F_{ext}\) is exerted to the left on the rightmost block, as shown in the figure. The masses of the blocks are indicated in the figure.
If the force is exerted for a time \( \Delta t \), what is the change in velocity of the center of mass of the three-block system?
Correct Answer: \( \boxed{\mathrm{A}} \)
The total mass of the three-block system is
\( m_{\text{sys}} = m + 3m + 2m = 6m \)
The only external horizontal force acting on the system is \(F_{ext}\).
Applying Newton’s Second Law to the entire system,
\( F_{ext} = m_{\text{sys}} a_{\text{CM}} \)
Therefore,
\( a_{\text{CM}} = \dfrac{F_{ext}}{6m} \)
Using the definition of acceleration,
\( a_{\text{CM}}=\dfrac{\Delta v_{\text{CM}}}{\Delta t} \)
Solving for the change in velocity of the center of mass:
\( \Delta v_{\text{CM}} = a_{\text{CM}}\Delta t \)
\( \Delta v_{\text{CM}} = \dfrac{F_{ext}}{6m}\Delta t \)
\( \Delta v_{\text{CM}} = \dfrac{F_{ext}\Delta t}{6m} \)
Thus, the change in velocity of the center of mass is
\( \boxed{\dfrac{F_{ext}\Delta t}{6m}} \).
Equivalently, using the impulse-momentum theorem for the entire system,
\( F_{ext}\Delta t = m_{\text{sys}}\Delta v_{\text{CM}} \),
which gives the same result.
A spaceship and its shuttle pod are traveling to the right in a straight line with speed \(v\), as shown in the top figure above. The mass of the pod is \(m\), and the mass of the spaceship is \(6m\). The pod is launched, and afterward the pod is moving to the right with speed \(v_p\) and the spaceship is moving to the right with speed \(v_f\), where \(v_f>v\), as shown in the bottom figure.
Which of the following is true of the speed \(v_c\) of the center of mass of the system after the pod is launched?
Correct Answer: \( \boxed{\mathrm{D}} \)
The launch of the pod is caused by internal forces within the spaceship-pod system.
Since no external force acts on the system in the horizontal direction, the total momentum of the system is conserved.
The velocity of the center of mass is given by
\( v_c=\dfrac{\sum mv}{\sum m} \)
Initially, both the spaceship and pod move together with speed \(v\), so
\( v_{c,i}=v \)
Because the total momentum remains constant and the total mass does not change, the velocity of the center of mass must remain constant.
Therefore,
\( v_{c,f}=v_{c,i}=v \)
Even though the spaceship speeds up to \(v_f\) and the pod moves at a different speed \(v_p\), the center of mass continues to move at its original speed.
Therefore, the speed of the center of mass after the pod is launched is
\( v_c=v \).
Therefore, the correct answer is (D).
