Consider two blocks $A$ and $B$ of masses $m_{1}=10 \mathrm{~kg}$ and $m_{2}=5 \mathrm{~kg}$ that ar

Question

Consider two blocks $A$ and $B$ of masses $m_{1}=10 \mathrm{~kg}$ and $m_{2}=5 \mathrm{~kg}$ that are placed on a frictionless table. The block $A$ moves with a constant speed $v=3 \mathrm{~m} / \mathrm{s}$ towards the block $B$ kept at rest. A spring with spring constant $k=3000 \mathrm{~N} / \mathrm{m}$ is attached with the block $B$ as shown in the figure. After the collision, suppose that the blocks $A$ and $B$, along with the spring in constant compression state, move together, then the compression in the spring is (Neglect the mass of the spring)\

Professor PC Thomas Classes · Answer

(c) : $v_{1}=v=3 \mathrm{~m} / \mathrm{s}$ Let the common velocity be $v_{\mathrm{cm}}$.\ By conservation of momentum\ $m_{1} v_{1}+m_{2} v_{2}=\left(m_{1}+m_{2}ight) v_{\mathrm{cm}}$\ $v_{\mathrm{cm}}=\frac{10 	imes 3+5 	imes 0}{15}=2 \mathrm{~m} / \mathrm{s}$\ Now, by conservation of energy\ $\frac{1}{2} k x^{2}=\frac{1}{2} 	imes 10 	imes 3^{2}-\left[\frac{1}{2} 	imes 15 	imes 2^{2}ight]$\ $\frac{1}{2} 	imes 3000 \cdot x^{2}=45-30 ; x^{2}=\frac{1}{100}$\ $x=0.1 \mathrm{~m}$\