Two bodies $A$ and $B$ of masses $m$ and $2 m$ respectively placed on a smooth floor are connected by a spring. A third body $C$ of mass $m$ moves with velocity $v_{0}$ along the line joining $A$ and $B$ and collides elastically with $A$ . At a certain instant of time $t_{0}$ after collision it is found that the instantaneous velocities of $A$ and $B$ are same. The compression in the spring at $t_{0}$ is $x_{0}$ , then:

The common velocity of

A

and

B

at time

t_{0}

\frac{v_{0}}{3}

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The spring constant is

k = \frac{3 m v_{0}^{2}}{2 x_{0}^{2}}

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The spring constant is

k = \frac{2 m v_{0}^{2}}{3 x_{0}^{2}}

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None

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Solution

The correct options are
A The common velocity of $A$ and $B$ at time $t_{0}$ is $\frac{v_{0}}{3}$ .
C The spring constant is $k = \frac{2 m v_{0}^{2}}{3 x_{0}^{2}}$ .
Applying momentum conservation law,
$p_{i} = p_{f}$
$3 m v^{^{'}} = m v_{0}$
$∴ v^{^{'}} = \frac{v_{0}}{3}$
Now,
$\frac{1}{2} m v_{0}^{2} = \frac{1}{2} m {(\frac{v_{0}}{3})}^{2} + \frac{1}{2} (2 m) {(\frac{v_{0}}{3})}^{2} + \frac{1}{2} k x_{0}^{2}$
$= \frac{1}{2} \frac{m v_{0}^{2}}{9} + \frac{1}{2} 2 m \frac{v_{0}^{2}}{9} + \frac{1}{2} k x_{0}^{2}$
$\Rightarrow m v_{0}^{2} = \frac{m v_{0}^{2}}{3} + k x_{0}^{2}$
$∴ k x_{0}^{2} = \frac{2}{3} m v_{0}^{2}$ $∴ k = \frac{2 m v_{0}^{2}}{3 x_{0}^{2}}$

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Similar questions

Q. Two bodies A and B of masses m and 2m respectively are placed on a smooth floor.They are connected by a spring .A third body C of mass m moves with velocity

v_{0}

along the line joining A and B and collides elastically with A as shown in fig.At a certain instant of time

t_{0}

after collision , it is found that the instantaneous velocities of A and B are the same.Further at this instant the compression of the spring is found to be

x_{0}

.Determine -(i) the common velocity of A and B at time

t_{0}

and (ii) the spring constant.

Two bodies A and B of mases m and 2m respectively are placed on a smooth floor. They are connected by a spring. A third body C of mass m moves with velocity $v_{0}$ along the line joining A and B and collides elastically with A as shown in figure. At a certain instant of time $t_{0}$ after collision, it is found that the instantaneous velocities of A and B are the same. Further at this instant the compression of the spring is found to be $x_{0}$ . Determine (a) the common velocity of A and B at time $t_{0}$ and (b) the spring constant.

Q. Two blocks A and B of masses m and 2m respectively placed on a smooth floor are connected by a spring. A third body C of mass m moves with velocity

v_{0}

along the line joining A and B and collides elastically with A. At a certain instant of time after collision it is found that the instantaneous velocities of A and B are same then :

Q. A block C of mass m is moving with velocity

v_{0}

and collides elastically with block A of mass m and connected to another block B of mass 2m through spring with spring constant K. What is the value of K, if

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The correct options are A The common velocity of A and B at time t0 is v03. C The spring constant is k=2mv203x20 .Applying momentum conservation law, pi=pf 3mv′=mv0 ∴v′=v03 Now, 12mv20=12m(v03)2+12(2m)(v03)2+12kx20 =12mv209+122mv209+12kx20 ⇒mv20=mv203+kx20 ∴kx20=23mv20 ∴k=2mv203x20