Question

Two blocks of masses $3 \,kg$ and $6\, kg$ rest on a horizontal smooth surface. The $3 \,kg$ block is attached to a spring with a force constant $k=900 \,Nm ^{-1}$ which is compressed $2\, m$ from the equilibrium position as shown in figure. The $6 \,kg$ mass is at rest at $1\, m$ from mean position. $3\, kg$ mass strikes the $6\, kg$ mass and the two stick together :

• A. Velocity of the combined masses immediately after the collision is $10\, ms ^{-1}$
• B. Velocity of the combined masses immediately after the collision is $5 \,ms ^{-1}$
• C. Amplitude of the resulting oscillation is $\sqrt{2} \,m$
• D. Amplitude of the resulting oscillation is $\sqrt{5 / 2} \,m$.

Answer 1

Answer: (A), (C)

When $3 \,kg$ mass is released the amplitude of its oscillations is $2 \,m$ and at a distance $1 \,m$ from the equilibrium position we can find the speed of it using the relation $v=\left[(k / m)\left(A^{2}-x^{2}\right)\right]^{1 / 2}$ then by conservation of momentum we can find the resulting speed of the combined mass and the new amplitude using the above relation