Question

A mass $M_1$ slides on a $45^\circ$ smooth inclined plane of height $H$ as shown. The mass $M_2$ moves downwards, strikes the ground and sticks to it. Masses of blocks are equal. The distance moved by mass $M_1$ along the inclined plane before it finally stops is $S$. Which of the following is correct? (Take: $\sqrt{2} = 1.4$).

• A. S = 0.8 H
• B. S = 0.6 H
• C. S = 0.4 H
• D. S = H

Answer 1

Answer: (B)

S = (√2 + 1) H/4

Answer 2

Let the string length change be $d$ (with $d = H/2$ so that $M_2$ hits the floor). Using energy (or kinematics) one finds that the common speed at that moment is $v$ with $v^2 = gH(1 – 1/√2)/2$.

After $M_2$ sticks, $M_1$ ascends an extra distance $sₑ = v²√2/(2g) = H(√2 –1)/4$.

Thus the total distance along the plane is $S = H/2 + H(√2–1)/4 = H(√2+1)/4$.