# [SOLVED] Mechanics - Collisions?

• May 16th 2009, 07:12 AM
fardeen_gen
[SOLVED] Mechanics - Collisions?
$A$ is a fixed point at a height $H$ above a perfectly inelastic smooth horizontal plane. A light inextensible string of length $L(>H)$ has one end attached to $A$ and other is to a heavy particle. The particle is held at the level of $A$ with string just taut and released from rest. Find the height of the particle above the plane when it is next instantaneously at rest.
• May 17th 2009, 09:44 PM
Momentum and energy
Hello fardeen_gen
Quote:

Originally Posted by fardeen_gen
$A$ is a fixed point at a height $H$ above a perfectly inelastic smooth horizontal plane. A light inextensible string of length $L(>H)$ has one end attached to $A$ and other is to a heavy particle. The particle is held at the level of $A$ with string just taut and released from rest. Find the height of the particle above the plane when it is next instantaneously at rest.

Suppose that the string makes an angle $\theta$ with the horizontal at the moment when the particle strikes the table. Then:

$\sin\theta = \frac{H}{L}$ (1)

If the velocity just before the impact with the table is $v$, and the mass of the particle is $m$, then, using Conservation of Energy:

$\tfrac12mv^2 = mgH$

$\Rightarrow v^2 = 2gH$ (2)

The particle now strikes the table, which is smooth and inelastic. So the impulse on the particle is vertical and the vertical component of the particle's velocity after impact is zero. Its horizontal component of velocity before impact is $v\sin\theta$. So, if the velocity along the table after impact is $v_1$, since the particle's horizontal momentum is conserved:

$v_1 = v\sin\theta$ (3)

The particle now slides along the table until the string once again becomes taut, where it once more makes an angle $\theta$ with the horizontal. As the string becomes taut it exerts an impulse along its length on the particle. So the particle's momentum perpendicular to the string is now conserved. Its velocity in this direction is $v_1\sin\theta$ and this is therefore the velocity of the particle after the string has become taut.

So, if the particle now rises to a height $h$ above the table, using Conservation of Energy:

$mgh = \tfrac12mv_1^2\sin^2\theta$

$=\tfrac12mv^2\sin^4\theta$, from (3)

$= \tfrac12m(2gH)\Big(\frac{H}{L}\Big)^4$, from (1) and (2)

$\Rightarrow h = \frac{H^5}{L^4}$