Name: Linear Momentum of a System of Particles
Uploaded: 2019-07-19
Duration: 10 min 32 s
Description: Linear momentum is one of the most important concepts in physics. Watch this video to dive deep to visualise linear momentum of a system of particles.

Question

Two balls having masses $m$ and $2 m$ are fastened to two light strings of the same length $l$ . The other ends of the strings are fixed at $O$ . The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic.

$(a)$ Find the velocities of the balls just after their collision.
$(b)$ How high will the balls rise after the collision?

Accepted Answer

Mass of the ball $A = m$

Mass of the ball $B = 2 m$

When the ball reaches the lower end, let the velocity of $m$ be $V_{A}$ and the velocity of $2 m$ be $V_{B}$ .

Here only force due to gravity is doing work so,

Using the work-energy theorem, total change in kinetic energy will be equal to work done by gravity. So,

$\frac{1}{2} \times m {V_{A}}^{2} - \frac{1}{2} \times m \times 0^{2} = m g l$

$\Rightarrow V_{A} = \sqrt{2 g l} . . . (i)$

Similarly, the velocity of the block having mass $2 m$ will be,

$V_{B} = \sqrt{2 g l} . . . (ii)$

$V_{A} = V_{B} = V . . . (iii)$

Let the velocities after collision be ${V_{A}}^{'}$ and ${V_{B}}^{'}$ , respectively.

Then coefficient of restitution $(a)$ ,

$\Rightarrow V_{A}' - V_{B}' = - 2 V . . . (i v)$

Since there is no force in horizontal direction during collision, so momentum in horizontal direction will be conserved during collision.

Using the law of conservation of momentum,

$m \times V - 2 m V = m {V_{A}}^{'} + 2 m {V_{B}}^{'}$

$\Rightarrow {V_{A}}^{'} + 2 {V_{B}}^{'} = - V . . . (v)$

Subtracting equations $(iv)$ from $(v)$ , we get,

$3 {V_{B}}^{'} = V$

$\Rightarrow {V_{B}}^{'} = \frac{V}{3} = \frac{\sqrt{2 g l}}{3}$

Substituting in equation $(v)$ , we get,

${V_{A}}^{'} - {V_{B}}^{'} = - 2 V$

$\Rightarrow {V_{A}}^{'} = - 2 V + {V_{B}}^{'}$

$= - 2 V + \frac{V}{3}, [∵ V_{B} = \sqrt{2 g l}]$

$= - \frac{\sqrt{50 g l}}{3}$

$(b)$ Let the height reached by the ball of mass $2 m$ and $m$ be $h_{B}$ and $h_{A}$ , respectively.

Using the work-energy theorem, we get,

For ball $B$ ,

$\frac{1}{2} \times 2 m \times 0^{2} - \frac{1}{2} \times 2 m \times {V^{'}}_{B}^{2} = - 2 m g h$

$\Rightarrow h_{B} = \frac{l}{9}$

Similarly, for height of $A$ ,

$\frac{1}{2} \times m \times 0^{2} - \frac{1}{2} \times m \times {({V^{'}}_{A})}^{2} = m g h_{2}$

$\Rightarrow h_{A} = \frac{25 l}{9}$

As $h_{A}$ is more than $2 l$ , the maximum height possible without breaking the string is $2 l$ , i.e., the lighter ball will move in a vertical circle.

So, the mass $m$ will rise by a distance $2 l$ .

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