Unit 2: Force and Translational Dynamics
Weightage : 20-15%
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 repeatedly 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?
Correct Answer: \( \boxed{\mathrm{E}} \)
The spring force acts only between the two blocks, so it is an internal force of the system.
Since the surface is frictionless, there is no external horizontal force acting on the two-block system.
Applying Newton’s second law to the center of mass,
\(\sum F_{\mathrm{ext}}=M_{\mathrm{total}}a_{\mathrm{CM}}\)
Because \(\sum F_{\mathrm{ext}}=0\),
\(a_{\mathrm{CM}}=0\)
Therefore, the velocity and speed of the center of mass remain constant throughout the entire motion, even though the spring continuously compresses and expands.
The internal spring force changes the individual velocities of the blocks but cannot change the motion of the system’s center of mass.
Therefore, the correct answer is (E).
A moon of mass \(m\) orbits a planet of mass \(49m\) in an elliptical orbit as shown above. When the moon is at point A, its distance from the center of the planet is \(r_A\) and its speed is \(v_0\). When the moon is at point B, its speed is \(5v_0\).
When the moon is at point A, the distance from the moon to the center of mass of the planet-moon system is most nearly
Correct Answer: \( \boxed{\mathrm{E}} \)
For a two-body system, the center of mass lies on the line joining the two masses and is given by
\(r_{\mathrm{CM}}=\dfrac{m_2}{m_1+m_2}d\)
where \(d\) is the separation between the masses.
Here, the planet has mass \(49m\), the moon has mass \(m\), and their separation is \(r_A\).
The distance from the moon to the center of mass is
\(d_{\text{moon}\rightarrow\mathrm{CM}}=\dfrac{49m}{49m+m}\,r_A\)
\(=\dfrac{49}{50}r_A\)
Since the planet is much more massive than the moon, the center of mass lies very close to the planet’s center and therefore almost the entire separation \(r_A\) is measured from the moon to the center of mass.
Therefore, the correct answer is (E).
An arrow of mass \(m\) and speed \(v_0\) strikes and sticks to one end of a meterstick of mass \(M\), as shown in the diagram above. The meterstick is initially at rest on a horizontal surface and free to move without friction. The speed of the center of mass of the stick-arrow system after the arrow strikes is given by which of the following expressions?
Correct Answer: \( \boxed{\mathrm{C}} \)
Since there are no external horizontal forces acting on the arrow-meterstick system, the total linear momentum is conserved. The speed of the center of mass is given by the total momentum divided by the total mass.
Initial momentum:
\(p_{\mathrm{i}}=mv_0\)
Total mass after the collision:
\(M_{\mathrm{total}}=M+m\)
Therefore,
\(v_{\mathrm{CM}}=\dfrac{p_{\mathrm{total}}}{M_{\mathrm{total}}}=\dfrac{mv_0}{M+m}\)
The location where the arrow strikes the meterstick affects the rotational motion of the system but does not affect the translational speed of its center of mass.
Therefore, the correct answer is (C).
