Question

The masses of blocks A and B are m and M respectively. Between A and B, there is a constant frictional force F and B can slide on a smooth horizontal surface. A is set in motion with velocity v₀ while B is at rest. What is the distance moved by A relative to B before they move with the same velocity? 

• A. \( \dfrac{mMv_0^2}{F(m-M)} \)
• B. \( \dfrac{mMv_0^2}{2F(m-M)} \)
• C. \( \dfrac{mMv_0^2}{F(m+M)} \)
• D. (mMv₀²)/(2F(m+M))

Hint:

Use relative motion when friction acts between blocks.

Answer 1

The Correct Option is D

Step 1: Relative acceleration: a = F((1)/(m) + (1)/(M))
Step 2: Relative velocity becomes zero: 0 = v₀² - 2as
Step 3: Solving for distance: s = (mMv₀²)/(2F(m+M))