Question

The masses of blocks \(A\) and \(B\) are \(m\) and \(M\) respectively. Between \(A\) and \(B\) there is a constant frictional force \(F\). Block \(B\) can slide on a smooth horizontal surface. \(A\) is set in motion with velocity \(v_0\) 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. \(\dfrac{mMv_0^2}{2F(m+M)}\)

Hint:

For friction between blocks: \[ a_{\text{rel}} = a_1 - a_2 \] Always use relative motion equations.

Answer 1

The Correct Option is D

Step 1: Accelerations: \[ a_A = -\frac{F}{m}, \quad a_B = \frac{F}{M} \]
Step 2: Relative acceleration: \[ a_{\text{rel}} = a_A - a_B = -F\left(\frac{1}{m}+\frac{1}{M}\right) \]
Step 3: Final relative velocity is zero.
Step 4: Using kinematics: \[ 0 = v_0^2 + 2a_{\text{rel}}x \]
Step 5: Solving: \[ x = \frac{mMv_0^2}{2F(m+M)} \]