You visited us 1 times! Enjoying our articles? Unlock Full Access!

Velocity of COM

Two blocks of...

Question

Two blocks of masses $m_{1}$ and $m_{2}$ are connected by a spring of constant k (given figure).The block of mass $m_{2}$ is given a sharp impulse so that it acquires a velocity $v_{0}$ towards right.Find (a) the velocity of the centre of mass,(b) the maximum elongation that the spring will suffer.

Open in App

Solution

(a)We know that,
Velocity of centre of mass $= \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

Here, $v_{1} = 0, v_{2} = v_{0}$
So we can write
Velocity of COM $= \frac{m_{1} \times 0 + m_{2} \times v_{0}}{m_{1} + m_{2}}$
Velocity of center of mass $= \frac{m_{2} v_{0}}{m_{1} + m_{2}}$

(b)
The spring will attain maximum elongation when both velocities of two blocks will attain are the velocity of the center of mass.

$x \to$ maximum elongation of spring.

Change in kinetic energy=potential energy stored in spring

$\Rightarrow \frac{1}{2} m_{2} v_{0}^{2} - \frac{1}{2} (m_{1} + m_{2}) {(\frac{m_{2} v_{0}}{m_{1} + m_{2}})}^{2}$

$= \frac{1}{2} k x^{2}$

$\Rightarrow m_{2} v_{0}^{2} (1 - \frac{m_{2}}{m_{1} + m_{2}}) = k x^{2}$

After solving
Elongation is
$x = {[\frac{m_{1} m_{2}}{K (m_{1} + m_{2})}]}^{\frac{1}{2}} V_{0}$

Suggest Corrections

Similar questions

m_{1}

m_{2}

k

m_{2}

v_{0}

m_{1}

m_{2}

k

m_{2}

v_{o}

m_{1}

m_{2}

m_{2}

V_{o}

m_{2}

v_{0}

Join BYJU'S Learning Program