Question

An impulse gives a velocity of $14\text{m/s}$ to the heavier block. Choose the correct statement(s):

• A. Velocity of C.O.M of the system of blocks is $10\text{m/s}$
• B. Maximum compression in the spring is $2\sqrt{14}\text{m}$
• C. Velocity of C.O.M of the system of blocks is $5\text{m/s}$
• D. Maximum compression in the spring is $\sqrt{14}\text{m}$

Answer 1

Answer: (A), (B)

Maximum compression in the spring is $2\sqrt{14}\text{m}$

At maximum compression, speed of both blocks will be equal and it will be equal to the speed of C.O.M.
Since no external forces on the system along horizontal, the linear momentum will remain conserved.
$P=m_{\text{system}}{v}_{\text{C.O.M}}$
Let the maximum compression in the spring be $x_0$ and speed of the C.O.M. be $v_0$.


$v_0=v_{\text{C.O.M}}=\frac{10\times 14+4\times 0}{10+4}$
$[\because v_{\text{C.O.M}}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}]$
$\Rightarrow v_{\text{C.O.M}}=v_0=10\text{m/s}...\left(1\right)$
Since there is no non-conservative forces acting, we can apply Law of conservation of mechanical energy.
$M.E_i=M.E_f$
$\Rightarrow K{E}_i+P{E}_i=K{E}_f+P{E}_f...\left(2\right)$
Here, $PE$ represents the elastic potential energy of spring, $PE=\frac{1}{2}k{x}^2$
Substituting in Eq.$\left(2\right)$ we get,
$\Rightarrow \frac{1}{2}\times 10\times {14}^2+0=\frac{1}{2}\times 10\times v_0^2+\frac{1}{2}\times 4\times v_0^2+\frac{1}{2}\times 10\times x_0^2$
$\Rightarrow 10\times {14}^2=10\times {10}^2+4\times {10}^2+10\times x_0^2$
$\because [v_0=10\text{m/s}]$
$\Rightarrow {14}^2={10}^2+4\times 10+x_0^2$
$\Rightarrow x_0=\sqrt{196-100-40}=\sqrt{56}$
$\therefore x_0=2\sqrt{14}\text{m}$
Option (a) and (b) are correct.