Question

Two particles P and Q of mass 20 g and 40 g respectively are projected simultaneously from point A and B with same velocity of 49 ms-1 making angles 45° and 135° with horizontal (figure). Distance between A and B is 245 m. Both the particles move in same plane. After collision particle P retraces its own path.

(i) Find the position of Q when it strikes the ground.

(ii) How much time is taken by Q in reaching the ground after collision?

Answer 1

In horizontal direction,

Velocity of particle P = 49 cos 45° = \(\frac{49}{\sqrt{2}}\ ms^{-1}\) and

Velocity of particle Q = 49 cos 135° = \(-\frac{49}{\sqrt{2}}\ ms^{-1}\)

Thus the speed of both particles in horizontal direction is equal and in opposite direction. Therefore both will collide above middle point of AB.

∴ Horizontal distance covered by both the particles \(=\frac{245}{2}m\)

∵ Range R = \(\frac{u^2\sin2\theta}{g}\)

\( = \frac{49\times49\times\sin90^\circ}{9.8}\)

= 245 m
Therefore, both particles will collide at highest point of their path where vertical velocity becomes zero.

∴ Before collision the velocity of P, v_P = \(\frac{49}{\sqrt{2}}\ ms^{-1}\)

and that of Q, v_Q = \(-\frac{49}{\sqrt{2}}ms^{-1}\)

After collision v'_P = \(-\frac{49}{\sqrt{2}}\ ms^{-1}\)

and v'_Q = ?

According to principle of conservation of momentum,

\((20\times10^{-3}) \times \frac{49}{\sqrt{2}} + (40\times10^{-3})\times(-\frac{49}{\sqrt{2}}) \)

\(=(20\times10^{-3})(-\frac{49}{\sqrt{2}}) + (40\times10^{-3})\times v'_Q\)

1 - 2 = -1 + 2 v'_Q

or 2 - 2 = 2 v'_Q

or v'_Q = 0

Therefore particle Q will fall freely after collision.

∵ Maximum height

\(H = \frac{u^2\sin^2\theta}{2g} = \frac{(49)^2\times\frac{1}{2}}{2\times9.8} = 61.25\ m\)

∴ Time taken by Q in falling upto ground

\(y = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2\times61.25}{9.8}}\)

or t = 3.53 s