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 $k=900N{m}^{-1}$ which is compressed 2m from beyond the equilibrium position. The 6 kg mass is at rest at 1m from mean position. 3kg mass strikes the 6kg mass and the two stick together.

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

Answer 1

Answer: (A), (C)

The correct options are
A

Velocity of the combined masses immediately after the collision is $10m{s}^{-1}$

C

Amplitude of the resulting oscillation is $\sqrt{2}m$

${\omega }_1=\sqrt{\frac{K}{m_1}}=\sqrt{\frac{900}{3}}=10\sqrt{3}\frac{rad}{s}{V}_1={\omega }_1\sqrt{A_1^2-x^2}=\left(10\sqrt{3}\right)\sqrt{(2{)}^2-(1{)}^2}=30\frac{m}{s}$
= velocity of 3 kg just before collision.
From momentum conservation,
$P_i=P_{f}3\left(30\right)=(3+6)v_2$
= velocity of combined mass just after collision
${\omega }_2=\sqrt{\frac{k}{m_1+m_2}}=\sqrt{\frac{900}{3+6}}=10\frac{rad}{s}$
Now, $v_2={\omega }_2\sqrt{A_2^2-x^2}$
$or10\sqrt{A_2^2-(1{)}^2}\therefore A_2=\sqrt{2}m$