Question

Two smooth spheres of the same radius, but which have different masses $m_{1} & m_{2}$ collide inelastically. Their velocities before collision are $13 m s^{- 1} & 5 m s^{- 1}$ respectively along the directions shown in the figure in which $cotθ = \frac{5}{12} .$ An observer $S'$ moving parallel to the positive y - axis with a constant speed of $5 m s^{- 1}$ observes this collision. He finds the final velocity of $m_{1}$ to be $5 m s^{- 1}$ along the y - direction and the total loss in the kinetic energy of the system to be $\frac{1}{72}$ of its initial value. Determine;

$(a)$ The ratio of the masses

$(b)$ The velocity of $m_{2}$ after collision with (respect to observer)

Accepted Answer

Let the velocity of mass $m_{2}$ after the collision is $v = (v_{x}, v_{y})$ and the ratio of mass is $\frac{m_{1}}{m_{2}} = x$ , then,

From conservation of momentum,

Along horizontal direction,

$m_{1} (13 \sin θ) = m_{2} v_{x}$

$\frac{m_{1}}{m_{2}} \times 13 \times \frac{12}{13} = v_{x}$

$v_{x} = 12 x . . . (1)$

Alon vertical direction,

$m_{1} \times (13 \cos θ) + m_{2} \times 5 = m_{1} \times (5 + 5) + m_{2} v_{y}$ ; here the final velocity of $m_{1}$ will be $m_{1} & m_{2}$ ; w.r.t. ground.

$\Rightarrow v_{y} = 5 - 5 x . . . (2)$

Now, the loss in kinetic energy is $\frac{1}{72}$ times in the observer'sfrom, so the vertical velocity will change since the observer is moving along $y$ -axis with speed $5 m s^{- 1}$ , so the initial kinetic energy of the particles in observer frame will be,

$K E_{i} = \frac{1}{2} m_{1} [{(13 \sin θ)}^{2} + {(13 \cos θ - 5)}^{2}] + \frac{1}{2} m_{2} {(5 - 5)}^{2}$

$K E_{i} = \frac{1}{2} m_{1} [12^{2} + 0^{2}] = 72 m_{1} . . . (3)$

The final kinetic energy of the system will be,

$K E_{f} = \frac{1}{2} m_{1} {(5)}^{2} + \frac{1}{2} m_{2} [v_{x}^{2} + {(v_{y} - 5)}^{2}]$

From equation $(1)$ and $(2)$ ,

$K E_{f} = \frac{25}{2} m_{1} + \frac{1}{2} m_{2} [{(12 x)}^{2} + {(- 5 x)}^{2}]$

$K E_{f} = \frac{25}{2} m_{1} + \frac{1}{2} m_{2} (169 x^{2}) . . . (4)$

Loss in kinetic energy will be given as,

$K E_{i} - K E_{f} = \frac{1}{72} \times K E_{i}$

From equation $(3)$ and $(4)$ ,

$72 m_{1} - [\frac{25}{2} m_{1} + \frac{1}{2} m_{2} (169 x^{2})] = \frac{1}{72} \times 72 m_{1}$

Using $x = \frac{m_{1}}{m_{2}}$ ; we get,

$72 x - (\frac{25}{2} x + \frac{169}{2} x^{2}) = x$

$x = \frac{117}{169} = \frac{9}{13}$ ; means the ratio of masses is $\frac{m_{1}}{m_{2}} = \frac{9}{13}$ .

From equation $(1)$ and $(2)$ , the velocity of $m_{2}$ in observer frame will be,

$v_{x} = 12 \times \frac{9}{13} = \frac{108}{13} m s^{- 1}$

$v_{y} - 5 = - 5 \times \frac{9}{13} = - \frac{45}{13} m s^{- 1}$

Hence,

$v_{S'} = \sqrt{v_{x}^{2} + {(v_{y} - 5)}^{2}} = \frac{1}{13} \sqrt{108^{2} + {(- 45)}^{2}} = 9 m s^{- 1}$ .

Important Questions on Centre of Mass, Momentum and Collisions

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