Question

A circular portion of radius \(R_2\) has been removed from one edge of a circular disc of radius \(R_1\). The correct expression for the centre of mass for the remaining portion of the disc is

• A. \(-\dfrac{R_2^2}{R_1+R_2}\)
• B. \(-\dfrac{R_2^2}{R_1-R_2}\)
• C. \(\dfrac{R_2^2}{R_1+R_2}\)
• D. \(-\dfrac{R_1^2}{R_1+R_2}\)

Hint:

For bodies with a portion removed, use the negative mass method. Treat the removed part as having negative mass and apply the centre of mass formula: \[ x_{\text{cm}}=\frac{\sum m_ix_i}{\sum m_i}. \]

Answer 1

The Correct Option is A

Step 1: Decompose impact velocity into components.
u_x = 5 cos30° = 5√3/2, u_y (downward) = 5 sin30° = 5/2.

Step 2: Apply coefficient of restitution to vertical component only.
v_x unchanged = 5√3/2. v_y = e·u_y = 0.2 × 5/2 = 1/2 (upward).

Step 3: Compute resultant rebound speed.
v = √(v_x² + v_y²) = √(75/4 + 1/4) = √(76/4) = √19.

Step 4: Final Answer:
√19 m/s.